So every time when you hear the word 🗣️ — ∫ integration it's like everyone freaks out as if it's some kind of monster. Seriously, many people don't even like math and chapters like integration just seem way more difficult. Even those who enjoy math often find integration to be a real pain. Integration has all sorts of formulas, methods & formats and each requiring a different approach to solve. But in the end, we usually solve it using a formula.
But if we’re going to use a formula anyway, why waste time on different methods? Wouldn't it be great to have a single formula that gives the answer directly, without any extra steps? Just like in quadratic equations, instead of finding factors we can just easily use the quadratic formula (Sridharacharya Formula) x = − b ± b 2 − 4 a c 2 a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} x = 2 a − b ± b 2 − 4 a c
So, I decided to dive into this topic. I spent quite a bit of time messing around with computation engines for calculations. After some trial and error in manipulating and rearranging the equations, I derived four direct formulas for four different integration formats involving quadratic equation. With these, you can solve integration problems in just 15 to 30 seconds. And if you practice a bit, you might even be able to do it in 10 seconds flat!
1. Integration of 1 a x 2 + b x + c \frac{1}{{a{x^2} + bx + c}} a x 2 + b x + c 1
Standard approach:
Example 1. Evaluate the following integral: ∫ 1 9 x 2 + 6 x + 5 d x \displaystyle \int{{\frac{1}{{9{x^2} + 6x + 5}}dx}} ∫ 9 x 2 + 6 x + 5 1 d x
Solution: First, we start by factoring out the leading coefficient of the quadratic term:
∫ 1 9 x 2 + 6 x + 5 d x = ∫ 1 9 ( x 2 + 2 3 x + 5 9 ) d x \int{{\frac{1}{{9{x^2} + 6x + 5}}dx}} = \int{\frac{1}{9\left(x^2+\frac{2}{3}x+\frac{5}{9}\right)}dx} ∫ 9 x 2 + 6 x + 5 1 d x = ∫ 9 ( x 2 + 3 2 x + 9 5 ) 1 d x Next, we complete the square by adding and subtracting the square of half the coefficient of the linear term:
1 9 ∫ 1 x 2 + 2 3 x + 5 9 d x = 1 9 ∫ 1 x 2 + 2 3 x + 5 9 + ( 1 2 . 2 3 ) 2 − ( 1 2 . 2 3 ) 2 d x = 1 9 ∫ 1 x 2 + 2 3 x + ( 1 3 ) 2 + 5 9 − 1 9 d x = 1 9 ∫ 1 ( x + 1 3 ) 2 + 5 9 − 1 9 d x = 1 9 ∫ 1 ( x + 1 3 ) 2 + 4 9 d x = 1 9 ∫ 1 ( x + 1 3 ) 2 + ( 2 3 ) 2 d x = 1 9 ∫ 1 z 2 + w 2 d z w h e r e [ z = ( x + 1 3 ) a n d w = 2 3 ] \begin{align*}
& \frac{1}{9}\int{\frac{1}{x^2+\frac{2}{3}x+\frac{5}{9}}dx} = \frac{1}{9}\int{\frac{1}{x^2+\frac{2}{3}x+\frac{5}{9}+\left(\frac{1}{2}.\frac{2}{3}\right)^2-\left(\frac{1}{2}.\frac{2}{3}\right)^2}dx} \\
&= \frac{1}{9}\int{\frac{1}{x^2+\frac{2}{3}x+\left(\frac{1}{3}\right)^2+\frac{5}{9}-\frac{1}{9}}dx} = \frac{1}{9}\int{\frac{1}{\left(x+\frac{1}{3}\right)^2+\frac{5}{9}-\frac{1}{9}}dx} \\
&= \frac{1}{9}\int{\frac{1}{\left(x+\frac{1}{3}\right)^2+\frac{4}{9}}dx} = \frac{1}{9}\int{\frac{1}{\left(x+\frac{1}{3}\right)^2+\left(\frac{2}{3}\right)^2}dx}\\
&= \frac{1}{9}\int\frac{1}{z^2+w^2}dz \quad \quad where \left[z=\left(x+\frac{1}{3}\right) \quad and\ \quad w=\frac{2}{3}\right]
\end{align*} 9 1 ∫ x 2 + 3 2 x + 9 5 1 d x = 9 1 ∫ x 2 + 3 2 x + 9 5 + ( 2 1 . 3 2 ) 2 − ( 2 1 . 3 2 ) 2 1 d x = 9 1 ∫ x 2 + 3 2 x + ( 3 1 ) 2 + 9 5 − 9 1 1 d x = 9 1 ∫ ( x + 3 1 ) 2 + 9 5 − 9 1 1 d x = 9 1 ∫ ( x + 3 1 ) 2 + 9 4 1 d x = 9 1 ∫ ( x + 3 1 ) 2 + ( 3 2 ) 2 1 d x = 9 1 ∫ z 2 + w 2 1 d z w h ere [ z = ( x + 3 1 ) an d w = 3 2 ] Using the standard integral formula for a sum of squares, we have:
∫ 1 z 2 + w 2 d z = 1 w t a n − 1 ( z w ) + C \int\frac{1}{z^2+w^2}dz = \frac{1}{w}{tan}^{-1}\left(\frac{z}{w}\right)+C ∫ z 2 + w 2 1 d z = w 1 t an − 1 ( w z ) + C Substituting back z z z w w w
1 9 ∫ 1 z 2 + w 2 d z = 1 9 ( 1 ( 2 3 ) ) t a n − 1 ( x + 1 3 2 3 ) + C = 1 6 t a n − 1 ( 3 x + 1 2 ) + C \frac{1}{9}\int\frac{1}{z^2+w^2}dz = \frac{1}{9}\left(\frac{1}{\left(\frac{2}{3}\right)}\right){tan}^{-1}\left(\frac{x+\frac{1}{3}}{\frac{2}{3}}\right)+C = \frac{1}{6}{tan}^{-1}\left(\frac{3x+1}{2}\right)+C 9 1 ∫ z 2 + w 2 1 d z = 9 1 ( ( 3 2 ) 1 ) t an − 1 ( 3 2 x + 3 1 ) + C = 6 1 t an − 1 ( 2 3 x + 1 ) + C Example 2. Evaluate the following integral: ∫ 1 2 x 2 + x − 1 d x \displaystyle \int{{\frac{1}{{2{x^2} + x - 1}}dx}} ∫ 2 x 2 + x − 1 1 d x
Solution: First, we start by factoring out the leading coefficient of the quadratic term:
∫ 1 2 x 2 + x − 1 d x = ∫ 1 2 ( x 2 + 1 2 x − 1 2 ) d x \int{{\frac{1}{{2{x^2} + x - 1}}dx}} = \int{\frac{1}{2\left(x^2+\frac{1}{2}x-\frac{1}{2}\right)}dx} ∫ 2 x 2 + x − 1 1 d x = ∫ 2 ( x 2 + 2 1 x − 2 1 ) 1 d x Next, we complete the square by adding and subtracting the square of half the coefficient of the linear term:
1 2 ∫ 1 x 2 + 1 2 x − 1 2 d x = 1 2 ∫ 1 x 2 + 1 2 x − 1 2 + ( 1 2 . 1 2 ) 2 − ( 1 2 . 1 2 ) 2 d x = 1 2 ∫ 1 x 2 + 1 2 x + ( 1 4 ) 2 − 1 2 − 1 16 d x = 1 2 ∫ 1 ( x + 1 4 ) 2 − 1 2 − 1 16 d x = 1 2 ∫ 1 ( x + 1 4 ) 2 − 9 16 d x = 1 2 ∫ 1 ( x + 1 4 ) 2 − ( 3 4 ) 2 d x = 1 2 ∫ 1 z 2 − w 2 d z w h e r e [ z = ( x + 1 4 ) a n d w = 3 4 ] \begin{align*}
& \frac{1}{2}\int{\frac{1}{x^2+\frac{1}{2}x-\frac{1}{2}}dx} = \frac{1}{2}\int{\frac{1}{x^2+\frac{1}{2}x-\frac{1}{2}+\left(\frac{1}{2}.\frac{1}{2}\right)^2-\left(\frac{1}{2}.\frac{1}{2}\right)^2}dx} \\
&= \frac{1}{2}\int{\frac{1}{x^2+\frac{1}{2}x+\left(\frac{1}{4}\right)^2-\frac{1}{2}-\frac{1}{16}}dx} = \frac{1}{2}\int{\frac{1}{\left(x+\frac{1}{4}\right)^2-\frac{1}{2}-\frac{1}{16}}dx} \\
&= \frac{1}{2}\int{\frac{1}{\left(x+\frac{1}{4}\right)^2-\frac{9}{16}}dx} = \frac{1}{2}\int{\frac{1}{\left(x+\frac{1}{4}\right)^2-\left(\frac{3}{4}\right)^2}dx} \\
&= \frac{1}{2}\int \frac{1}{z^2 - w^2} \ dz \quad \quad where \left[z=\left(x+\frac{1}{4}\right) \quad and \quad w=\frac{3}{4}\right]
\end{align*} 2 1 ∫ x 2 + 2 1 x − 2 1 1 d x = 2 1 ∫ x 2 + 2 1 x − 2 1 + ( 2 1 . 2 1 ) 2 − ( 2 1 . 2 1 ) 2 1 d x = 2 1 ∫ x 2 + 2 1 x + ( 4 1 ) 2 − 2 1 − 16 1 1 d x = 2 1 ∫ ( x + 4 1 ) 2 − 2 1 − 16 1 1 d x = 2 1 ∫ ( x + 4 1 ) 2 − 16 9 1 d x = 2 1 ∫ ( x + 4 1 ) 2 − ( 4 3 ) 2 1 d x = 2 1 ∫ z 2 − w 2 1 d z w h ere [ z = ( x + 4 1 ) an d w = 4 3 ] Using the known integral formula for a difference of squares, we have:
∫ 1 z 2 − w 2 d z = 1 2 w ln ∣ z − w z + w ∣ + C \int \frac{1}{z^2 - w^2} \ dz = \frac{1}{2w} \ln \left| \frac{z - w}{z + w} \right| + C ∫ z 2 − w 2 1 d z = 2 w 1 ln z + w z − w + C Substituting back z z z w w w
1 2 ∫ 1 z 2 − w 2 d z = 1 2 ( 1 2 ( 3 4 ) ) l n ∣ ( x + 1 4 ) − 3 4 ( x + 1 4 ) + 3 4 ∣ + C = 1 3 l n ∣ 2 x − 1 2 x + 2 ∣ + C \frac{1}{2}\int \frac{1}{z^2 - w^2} \ dz=\frac{1}{2}\left(\frac{1}{2\left(\frac{3}{4}\right)}\right){ln}\left|\frac{\left(x+\frac{1}{4}\right)-\frac{3}{4}}{\left(x+\frac{1}{4}\right)+\frac{3}{4}}\right|+C = \frac{1}{3}ln\left|\frac{2x-1}{2x+2}\right|+C 2 1 ∫ z 2 − w 2 1 d z = 2 1 ( 2 ( 4 3 ) 1 ) l n ( x + 4 1 ) + 4 3 ( x + 4 1 ) − 4 3 + C = 3 1 l n 2 x + 2 2 x − 1 + C However, this method can be a bit less effective because it requires lengthy algebraic manipulations to complete the square of the quadratic equation and you also need to remember the standard forms and integral formulas.
Simplified formula: Q = a x 2 + b x + c Q = ax^2 + bx + c Q = a x 2 + b x + c Q ′ = 2 a x + b Q' = 2ax + b Q ′ = 2 a x + b D = b 2 − 4 a c D = b^2 - 4ac D = b 2 − 4 a c
∫ 1 a x 2 + b x + c d x = ∫ 1 Q d x \int \frac{1}{ax^2 + bx + c} dx = \int \frac{1}{Q} dx ∫ a x 2 + b x + c 1 d x = ∫ Q 1 d x When ( D = b 2 − 4 a c ) < 0 : (D=b^2-4ac)<0: ( D = b 2 − 4 a c ) < 0 : —
To simplify the expression, we complete the square for the quadratic term. By adding and subtracting the square of half the coefficient of the linear term, i.e ( 1 2 . b a ) 2 \left(\frac{1}{2}.\frac{b}{a}\right)^2 ( 2 1 . a b ) 2
Q = a x 2 + b x + c = a ( x 2 + b a x + c a ) = a ( x 2 + b a x + c a + ( 1 2 . b a ) 2 − ( 1 2 . b a ) 2 ) Q = ax^2+bx+c = a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right) = a\left(x^2+\frac{b}{a}x+\frac{c}{a}+\left(\frac{1}{2}.\frac{b}{a}\right)^2-\left(\frac{1}{2}.\frac{b}{a}\right)^2\right) Q = a x 2 + b x + c = a ( x 2 + a b x + a c ) = a ( x 2 + a b x + a c + ( 2 1 . a b ) 2 − ( 2 1 . a b ) 2 ) = a ( ( x + b 2 a ) 2 + c a − ( b 2 a ) 2 ) = a ( ( x + b 2 a ) 2 + c a − b 2 4 a 2 ) = a ( ( x + b 2 a ) 2 + ( 4 a c − b 2 4 a 2 ) ) = a ( ( x + b 2 a ) 2 + ( 4 a c − b 2 4 a 2 ) 2 ) \begin{align*}
&= a\left({\left(x+\frac{b}{2a}\right)^2+\frac{c}{a}-\left(\frac{b}{2a}\right)^2}\right) = a\left({\left(x+\frac{b}{2a}\right)^2+\frac{c}{a}-\frac{b^2}{4a^2}}\right) \\
&= a\left({\left(x+\frac{b}{2a}\right)^2+\left(\frac{4ac-b^2}{4a^2}\right)}\right) = a\left({\left(x+\frac{b}{2a}\right)^2+\left(\sqrt\frac{4ac-b^2}{4a^2}\right)^2}\right)
\end{align*} = a ( ( x + 2 a b ) 2 + a c − ( 2 a b ) 2 ) = a ( ( x + 2 a b ) 2 + a c − 4 a 2 b 2 ) = a ( ( x + 2 a b ) 2 + ( 4 a 2 4 a c − b 2 ) ) = a ( x + 2 a b ) 2 + ( 4 a 2 4 a c − b 2 ) 2 Thus, the integral we want to solve simplifies to the following form:
∫ 1 Q d x = ∫ 1 a x 2 + b x + c d x = ∫ 1 a ( ( x + b 2 a ) 2 + ( 4 a c − b 2 4 a 2 ) 2 ) d x \int\frac{1}{Q}dx = \int\frac{1}{ax^2+bx+c}dx = \int\frac{1}{a\left({\left(x+\frac{b}{2a}\right)^2+\left(\sqrt\frac{4ac-b^2}{4a^2}\right)^2}\right)}dx ∫ Q 1 d x = ∫ a x 2 + b x + c 1 d x = ∫ a ( ( x + 2 a b ) 2 + ( 4 a 2 4 a c − b 2 ) 2 ) 1 d x N o w , ∫ 1 a ( ( x + b 2 a ) 2 + ( 4 a c − b 2 4 a 2 ) 2 ) d x = 1 a ∫ 1 ( x + b 2 a ) 2 + ( 4 a c − b 2 4 a 2 ) 2 d x Now, \quad \int\frac{1}{a\left({\left(x+\frac{b}{2a}\right)^2+\left(\sqrt\frac{4ac-b^2}{4a^2}\right)^2}\right)}dx = \frac{1}{a}\int\frac{1}{{\left(x+\frac{b}{2a}\right)^2+\left(\sqrt\frac{4ac-b^2}{4a^2}\right)^2}}dx N o w , ∫ a ( ( x + 2 a b ) 2 + ( 4 a 2 4 a c − b 2 ) 2 ) 1 d x = a 1 ∫ ( x + 2 a b ) 2 + ( 4 a 2 4 a c − b 2 ) 2 1 d x = 1 a ∫ 1 z 2 + w 2 d z w h e r e [ w = 4 a c − b 2 4 a 2 a n d z = ( x + b 2 a ) ] = \frac{1}{a}\int\frac{1}{z^2+w^2}dz \quad \quad where\left[w=\sqrt\frac{4ac-b^2}{4a^2} \quad and \quad z=\left(x+\frac{b}{2a}\right)\right] = a 1 ∫ z 2 + w 2 1 d z w h ere [ w = 4 a 2 4 a c − b 2 an d z = ( x + 2 a b ) ] Using the known integral formula for a sum of squares, we have:
∫ 1 z 2 + w 2 d z = 1 w t a n − 1 ( z w ) + C \int\frac{1}{z^2+w^2}dz = \frac{1}{w}{tan}^{-1}\left(\frac{z}{w}\right)+C ∫ z 2 + w 2 1 d z = w 1 t an − 1 ( w z ) + C Substituting back z z z w w w
1 a ∫ 1 z 2 + w 2 d z = 1 a ( 1 4 a c − b 2 4 a 2 t a n − 1 ( ( x + b 2 a ) 4 a c − b 2 4 a 2 ) ) + C \frac{1}{a}\int\frac{1}{z^2+w^2}dz = \frac{1}{a}\left(\frac{1}{\sqrt\frac{4ac-b^2}{4a^2}}{tan}^{-1}\left(\frac{\left(x+\frac{b}{2a}\right)}{\sqrt\frac{4ac-b^2}{4a^2}}\right)\right)+C a 1 ∫ z 2 + w 2 1 d z = a 1 4 a 2 4 a c − b 2 1 t an − 1 4 a 2 4 a c − b 2 ( x + 2 a b ) + C Simplifying this expression further, we arrive at the following result:
1 a ( 4 a 2 4 a c − b 2 t a n − 1 ( 2 a x + b 2 a 4 a c − b 2 4 a 2 ) ) + C = 1 a ( 2 a 4 a c − b 2 t a n − 1 ( 2 a x + b 4 a c − b 2 ) ) + C \frac{1}{a}\left(\frac{\sqrt{4a^2}}{\sqrt{4ac-b^2}}{tan}^{-1}\left(\frac{\frac{2ax+b}{2a}}{\frac{\sqrt{4ac-b^2}}{\sqrt{4a^2}}}\right)\right)+C = \frac{1}{a}\left(\frac{{2a}}{\sqrt{4ac-b^2}}{tan}^{-1}\left(\frac{{2ax+b}}{{\sqrt{4ac-b^2}}}\right)\right)+C a 1 4 a c − b 2 4 a 2 t an − 1 4 a 2 4 a c − b 2 2 a 2 a x + b + C = a 1 ( 4 a c − b 2 2 a t an − 1 ( 4 a c − b 2 2 a x + b ) ) + C = 1 a ( 2 a ∣ b 2 − 4 a c ∣ t a n − 1 ( 2 a x + b ∣ b 2 − 4 a c ∣ ) ) + C = 2 ∣ D ∣ tan − 1 ( Q ′ ∣ D ∣ ) + C = \frac{1}{a}\left(\frac{{2a}}{\sqrt{\left|b^2-4ac\right|}}{tan}^{-1}\left(\frac{{2ax+b}}{{\sqrt{\left|b^2-4ac\right|}}}\right)\right)+C = {\frac{2}{\sqrt{|D|}}{\tan^{-1}\left(\frac{Q'}{\sqrt{|D|}}\right)}} + C = a 1 ( ∣ b 2 − 4 a c ∣ 2 a t an − 1 ( ∣ b 2 − 4 a c ∣ 2 a x + b ) ) + C = ∣ D ∣ 2 tan − 1 ( ∣ D ∣ Q ′ ) + C Therefore, after solving we get the simplified formula for the integral of the quadratic equation:
w h e n D < 0 : ∫ 1 a x 2 + b x + c d x = ∫ 1 Q d x = 2 ∣ D ∣ tan − 1 ( Q ′ ∣ D ∣ ) + C when \quad D<0: \int \frac{1}{ax^2 + bx + c} dx = \int\frac{1}{Q}dx = {\frac{2}{\sqrt{|D|}}{\tan^{-1}\left(\frac{Q'}{\sqrt{|D|}}\right)}} + C w h e n D < 0 : ∫ a x 2 + b x + c 1 d x = ∫ Q 1 d x = ∣ D ∣ 2 tan − 1 ( ∣ D ∣ Q ′ ) + C When ( D = b 2 − 4 a c ) > 0 : (D=b^2-4ac)>0: ( D = b 2 − 4 a c ) > 0 : —
To simplify the expression, we complete the square for the quadratic term. By adding and subtracting the square of half the coefficient of the linear term, i.e ( 1 2 . b a ) 2 \left(\frac{1}{2}.\frac{b}{a}\right)^2 ( 2 1 . a b ) 2
Q = a x 2 + b x + c = a ( x 2 + b a x + c a ) = a ( x 2 + b a x + c a + ( 1 2 . b a ) 2 − ( 1 2 . b a ) 2 ) Q = ax^2+bx+c = a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right) = a\left(x^2+\frac{b}{a}x+\frac{c}{a}+\left(\frac{1}{2}.\frac{b}{a}\right)^2-\left(\frac{1}{2}.\frac{b}{a}\right)^2\right) Q = a x 2 + b x + c = a ( x 2 + a b x + a c ) = a ( x 2 + a b x + a c + ( 2 1 . a b ) 2 − ( 2 1 . a b ) 2 ) = a ( ( x + b 2 a ) 2 + c a − ( b 2 a ) 2 ) = a ( ( x + b 2 a ) 2 + c a − b 2 4 a 2 ) = a ( ( x + b 2 a ) 2 − ( b 2 − 4 a c 4 a 2 ) ) = a ( ( x + b 2 a ) 2 − ( b 2 − 4 a c 4 a 2 ) 2 ) \begin{align*}
&= a\left({\left(x+\frac{b}{2a}\right)^2+\frac{c}{a}-\left(\frac{b}{2a}\right)^2}\right) = a\left({\left(x+\frac{b}{2a}\right)^2+\frac{c}{a}-\frac{b^2}{4a^2}}\right) \\
&= a\left({\left(x+\frac{b}{2a}\right)^2-\left(\frac{b^2-4ac}{4a^2}\right)}\right) = a\left({\left(x+\frac{b}{2a}\right)^2-\left(\sqrt\frac{b^2-4ac}{4a^2}\right)^2}\right)
\end{align*} = a ( ( x + 2 a b ) 2 + a c − ( 2 a b ) 2 ) = a ( ( x + 2 a b ) 2 + a c − 4 a 2 b 2 ) = a ( ( x + 2 a b ) 2 − ( 4 a 2 b 2 − 4 a c ) ) = a ( x + 2 a b ) 2 − ( 4 a 2 b 2 − 4 a c ) 2 Thus, the integral we want to solve simplifies to the following form:
∫ 1 Q d x = ∫ 1 a x 2 + b x + c d x = ∫ 1 a ( ( x + b 2 a ) 2 − ( b 2 − 4 a c 4 a 2 ) 2 ) d x \int\frac{1}{Q}dx = \int\frac{1}{ax^2+bx+c}dx = \int\frac{1}{a\left({\left(x+\frac{b}{2a}\right)^2-\left(\sqrt\frac{b^2-4ac}{4a^2}\right)^2}\right)}dx ∫ Q 1 d x = ∫ a x 2 + b x + c 1 d x = ∫ a ( ( x + 2 a b ) 2 − ( 4 a 2 b 2 − 4 a c ) 2 ) 1 d x N o w , ∫ 1 a ( ( x + b 2 a ) 2 − ( b 2 − 4 a c 4 a 2 ) 2 ) d x = 1 a ∫ 1 ( x + b 2 a ) 2 − ( b 2 − 4 a c 4 a 2 ) 2 d x Now, \quad \int\frac{1}{a\left({\left(x+\frac{b}{2a}\right)^2-\left(\sqrt\frac{b^2-4ac}{4a^2}\right)^2}\right)}dx = \frac{1}{a}\int\frac{1}{{\left(x+\frac{b}{2a}\right)^2-\left(\sqrt\frac{b^2-4ac}{4a^2}\right)^2}}dx N o w , ∫ a ( ( x + 2 a b ) 2 − ( 4 a 2 b 2 − 4 a c ) 2 ) 1 d x = a 1 ∫ ( x + 2 a b ) 2 − ( 4 a 2 b 2 − 4 a c ) 2 1 d x = 1 a ∫ 1 z 2 − w 2 d z w h e r e [ w = b 2 − 4 a c 4 a 2 a n d z = ( x + b 2 a ) ] = \frac{1}{a}\int\frac{1}{z^2-w^2}dz \quad \quad where\left[w=\sqrt\frac{b^2-4ac}{4a^2} \quad and \quad z=\left(x+\frac{b}{2a}\right)\right] = a 1 ∫ z 2 − w 2 1 d z w h ere [ w = 4 a 2 b 2 − 4 a c an d z = ( x + 2 a b ) ] Using the known integral formula for a difference of squares, we have:
∫ 1 z 2 − w 2 d z = 1 2 w ln ∣ z − w z + w ∣ + C \int \frac{1}{z^2 - w^2} \ dz = \frac{1}{2w} \ln \left| \frac{z - w}{z + w} \right| + C ∫ z 2 − w 2 1 d z = 2 w 1 ln z + w z − w + C Substituting back z z z w w w
1 a ∫ 1 z 2 − w 2 d z = 1 a ( 1 2 ( b 2 − 4 a c 4 a 2 ) l n ∣ ( x + b 2 a ) − b 2 − 4 a c 4 a 2 ( x + b 2 a ) + b 2 − 4 a c 4 a 2 ∣ ) + C \frac{1}{a}\int\frac{1}{z^2-w^2}dz = \frac{1}{a}\left(\frac{1}{2\left(\sqrt\frac{b^2-4ac}{4a^2}\right)}ln\left|\frac{\left(x+\frac{b}{2a}\right)-\sqrt\frac{b^2-4ac}{4a^2}}{\left(x+\frac{b}{2a}\right)+\sqrt\frac{b^2-4ac}{4a^2}}\right|\right)+C a 1 ∫ z 2 − w 2 1 d z = a 1 2 ( 4 a 2 b 2 − 4 a c ) 1 l n ( x + 2 a b ) + 4 a 2 b 2 − 4 a c ( x + 2 a b ) − 4 a 2 b 2 − 4 a c + C Simplifying this expression further, we arrive at the following result:
1 a ( 4 a 2 2 ( b 2 − 4 a c ) l n ∣ 2 a x + b 2 a − b 2 − 4 a c 4 a 2 2 a x + b 2 a + b 2 − 4 a c 4 a 2 ∣ ) + C = 1 a ( 2 a 2 ( b 2 − 4 a c ) l n ∣ 2 a x + b 2 a − b 2 − 4 a c 2 a 2 a x + b 2 a + b 2 − 4 a c 2 a ∣ ) + C \frac{1}{a}\left(\frac{\sqrt{4a^2}}{2\left(\sqrt{b^2-4ac}\right)}ln\left|\frac{\frac{2ax+b}{2a}-\frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}}{\frac{2ax+b}{2a}+\frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}}\right|\right)+C = \frac{1}{a}\left(\frac{{2a}}{2\left(\sqrt{b^2-4ac}\right)}ln\left|\frac{\frac{2ax+b}{2a}-\frac{\sqrt{b^2-4ac}}{{2a}}}{\frac{2ax+b}{2a}+\frac{\sqrt{b^2-4ac}}{{2a}}}\right|\right)+C a 1 2 ( b 2 − 4 a c ) 4 a 2 l n 2 a 2 a x + b + 4 a 2 b 2 − 4 a c 2 a 2 a x + b − 4 a 2 b 2 − 4 a c + C = a 1 ( 2 ( b 2 − 4 a c ) 2 a l n 2 a 2 a x + b + 2 a b 2 − 4 a c 2 a 2 a x + b − 2 a b 2 − 4 a c ) + C = 1 b 2 − 4 a c l n ∣ 2 a x + b − b 2 − 4 a c 2 a 2 a x + b + b 2 − 4 a c 2 a ∣ + C = 1 b 2 − 4 a c l n ∣ 2 a x + b − b 2 − 4 a c 2 a x + b + b 2 − 4 a c ∣ + C = 1 D ln ∣ Q ′ − D Q ′ + D ∣ + C = \frac{{1}}{\sqrt{b^2-4ac}}ln\left|\frac{\frac{{2ax+b}-{\sqrt{b^2-4ac}}}{2a}}{\frac{{2ax+b}+{\sqrt{b^2-4ac}}}{2a}}\right|+C = \frac{{1}}{\sqrt{b^2-4ac}}ln\left|\frac{{2ax+b}-{\sqrt{b^2-4ac}}}{{2ax+b}+{\sqrt{b^2-4ac}}}\right|+C = \frac{1}{\sqrt{D}} \ln\left|\frac{{Q'} - \sqrt{D}}{{Q'} + \sqrt{{D}}}\right|+C = b 2 − 4 a c 1 l n 2 a 2 a x + b + b 2 − 4 a c 2 a 2 a x + b − b 2 − 4 a c + C = b 2 − 4 a c 1 l n 2 a x + b + b 2 − 4 a c 2 a x + b − b 2 − 4 a c + C = D 1 ln Q ′ + D Q ′ − D + C Therefore, after solving we get the simplified formula for the integral of the quadratic equation:
w h e n D > 0 & a > 0 : ∫ 1 a x 2 + b x + c d x = ∫ 1 Q d x = + 1 D ln ∣ Q ′ − D Q ′ + D ∣ + C when \quad D>0 \quad \& \quad a>0: \int\frac{1}{ax^2+bx+c}dx = \int\frac{1}{Q}dx = +\frac{1}{\sqrt{D}} \ln\left|\frac{{Q'} - \sqrt{D}}{{Q'} + \sqrt{{D}}}\right|+C w h e n D > 0 & a > 0 : ∫ a x 2 + b x + c 1 d x = ∫ Q 1 d x = + D 1 ln Q ′ + D Q ′ − D + C w h e n D > 0 & a < 0 : ∫ 1 a x 2 + b x + c d x = ∫ 1 Q d x = − 1 D ln ∣ Q ′ − D Q ′ + D ∣ + C when \quad D>0 \quad \& \quad a<0: \int\frac{1}{ax^2+bx+c}dx = \int\frac{1}{Q}dx = -\frac{1}{\sqrt{D}} \ln\left|\frac{{Q'} - \sqrt{D}}{{Q'} + \sqrt{{D}}}\right|+C w h e n D > 0 & a < 0 : ∫ a x 2 + b x + c 1 d x = ∫ Q 1 d x = − D 1 ln Q ′ + D Q ′ − D + C Alright, let's test out this formula with a question to see if it actually works and how much time it could save us. First, we'll solve it using the old methods everyone uses, and then we'll try out our new formula.
Question 1. Evaluate the following integral:
∫ 1 2 x 2 − 3 x + 2 d x \int{{\frac{1}{{2{x^2} - 3x + 2}}dx}} ∫ 2 x 2 − 3 x + 2 1 d x Using step by step method: u = 2 x 2 − 3 x + 2 u = 2{x^2} - 3x + 2 u = 2 x 2 − 3 x + 2
2 x 2 − 3 x + 2 = 2 ( x 2 − 3 2 x + 1 ) = 2 ( x 2 − 3 2 x + 9 16 − 9 16 + 1 ) = 2 ( ( x − 3 4 ) 2 + 7 16 ) \begin{align*}
2{x^2} - 3x + 2 &= 2\left( {x^2 - \frac{3}{2}x + 1} \right) \\
&= 2\left( {x^2 - \frac{3}{2}x + \frac{9}{16} - \frac{9}{16} + 1} \right) \\
&= 2\left( {\left( {x - \frac{3}{4}} \right)^2 + \frac{7}{16}} \right)
\end{align*} 2 x 2 − 3 x + 2 = 2 ( x 2 − 2 3 x + 1 ) = 2 ( x 2 − 2 3 x + 16 9 − 16 9 + 1 ) = 2 ( ( x − 4 3 ) 2 + 16 7 ) With this the integral is:
∫ 1 2 x 2 − 3 x + 2 d x = 1 2 ∫ 1 ( x − 3 4 ) 2 + 7 16 d x \int \frac{1}{2{x^2} - 3x + 2} dx = \frac{1}{2} \int \frac{1}{\left( {x - \frac{3}{4}} \right)^2 + \frac{7}{16}} dx ∫ 2 x 2 − 3 x + 2 1 d x = 2 1 ∫ ( x − 4 3 ) 2 + 16 7 1 d x Now this may not seem like all that great of a change. However, notice that we can now use the following substitution.
u = x − 3 4 ; d u = d x u = x - \frac{3}{4} \quad ; \quad du = dx u = x − 4 3 ; d u = d x and the integral is now:
∫ 1 2 x 2 − 3 x + 2 d x = 1 2 ∫ 1 u 2 + 7 16 d u \int \frac{1}{2{x^2} - 3x + 2} dx = \frac{1}{2} \int \frac{1}{u^2 + \frac{7}{16}} du ∫ 2 x 2 − 3 x + 2 1 d x = 2 1 ∫ u 2 + 16 7 1 d u We can now see that this is an inverse tangent! So, using the formula we get,
∫ 1 2 x 2 − 3 x + 2 d x = 1 2 ( 4 7 ) tan − 1 ( 4 u 7 ) + C = 2 7 tan − 1 ( 4 x − 3 7 ) + C \begin{align*}
\int \frac{1}{2{x^2} - 3x + 2} dx &= \frac{1}{2} \left( \frac{4}{\sqrt{7}} \right) \tan^{-1} \left( \frac{4u}{\sqrt{7}} \right) + C \\
&= \frac{2}{\sqrt{7}} \tan^{-1} \left( \frac{4x - 3}{\sqrt{7}} \right) + C
\end{align*} ∫ 2 x 2 − 3 x + 2 1 d x = 2 1 ( 7 4 ) tan − 1 ( 7 4 u ) + C = 7 2 tan − 1 ( 7 4 x − 3 ) + C Using the new formula: 2 x 2 − 3 x + 2 2{x^2} - 3x + 2 2 x 2 − 3 x + 2 Q Q Q Q ′ = 4 x − 3 Q' = 4x-3 Q ′ = 4 x − 3 D = b 2 − 4 a c D=b^2-4ac D = b 2 − 4 a c
On comparing the quadratic equation 2 x 2 − 3 x + 2 2{x^2} - 3x + 2 2 x 2 − 3 x + 2 a x 2 + b x + c ax^2+bx+c a x 2 + b x + c a = 2 , b = − 3 a=2, b=-3 a = 2 , b = − 3 c = 2 c=2 c = 2
D = b 2 − 4 a c = ( − 3 ) 2 − 4 ( 2 ) ( 2 ) = − 7 \begin{align*}
D &=b^2-4ac \\
&= (-3)^2-4(2)(2) \\
&= -7
\end{align*} D = b 2 − 4 a c = ( − 3 ) 2 − 4 ( 2 ) ( 2 ) = − 7 Now let's put the values of variables in the formula for D < 0 D<0 D < 0
∫ 1 Q d x = ∫ 1 2 x 2 − 3 x + 2 d x = 2 ∣ D ∣ tan − 1 ( Q ′ ∣ D ∣ ) + C = 2 7 tan − 1 ( 4 x − 3 7 ) + C \begin{align*}
\int \frac{1}{Q} dx &= \int \frac{1}{2{x^2} - 3x + 2} dx \\
&= {\frac{2}{\sqrt{|D|}}{\tan^{-1}\left(\frac{Q'}{\sqrt{|D|}}\right)}} + C \\
&= \frac{2}{\sqrt{7}} \tan^{-1} \left( \frac{4x - 3}{\sqrt{7}} \right) + C
\end{align*} ∫ Q 1 d x = ∫ 2 x 2 − 3 x + 2 1 d x = ∣ D ∣ 2 tan − 1 ( ∣ D ∣ Q ′ ) + C = 7 2 tan − 1 ( 7 4 x − 3 ) + C So, as you can see this formula really works and it's super easy to use :) Unlike the old step by step method, there aren't too many calculations and it takes much less time. You just need to find the value of discriminant and see if it's positive or negative. Then, differentiate the quadratic once and plug in the values. Where it would normally take around 1-2 minutes to solve the question step by step, with this formula you can get the answer in just 30 seconds :)
Alright, let's solve another question using the formula where the discriminant of the quadratic equation will be positive.
Question 2. Evaluate the following integral:
∫ 1 2 x 2 + x − 1 d x \int{{\frac{1}{{2{x^2} + x - 1}}dx}} ∫ 2 x 2 + x − 1 1 d x Again let's assume the quadratic 2 x 2 + x − 1 2{x^2} + x -1 2 x 2 + x − 1 Q Q Q Q ′ = 4 x + 1 Q' = 4x+1 Q ′ = 4 x + 1 D = b 2 − 4 a c D=b^2-4ac D = b 2 − 4 a c
On comparing the quadratic equation 2 x 2 + x − 1 2{x^2} + x -1 2 x 2 + x − 1 a x 2 + b x + c ax^2+bx+c a x 2 + b x + c a = 2 , b = 1 a=2, b=1 a = 2 , b = 1 c = − 1 c=-1 c = − 1
D = b 2 − 4 a c = ( 1 ) 2 − 4 ( 2 ) ( − 1 ) = 9 \begin{align*}
D &=b^2-4ac \\
&= (1)^2-4(2)(-1) \\
&= 9
\end{align*} D = b 2 − 4 a c = ( 1 ) 2 − 4 ( 2 ) ( − 1 ) = 9 Now let's put the values of variables in the formula for D > 0 D>0 D > 0
∫ 1 Q d x = ∫ 1 2 x 2 + x − 1 d x = 1 D ln ∣ Q ′ − D Q ′ + D ∣ + C = 1 9 ln ∣ 4 x + 1 − 9 4 x + 1 + 9 ∣ + C = 1 3 ln ∣ 2 x − 1 2 x + 2 ∣ + C \begin{align*}
\int \frac{1}{Q} dx &= \int \frac{1}{2{x^2} + x - 1} dx \\
&= \frac{1}{\sqrt{D}} \ln\left|\frac{{Q'} - \sqrt{D}}{{Q'} + \sqrt{{D}}}\right|+C \\
&= \frac{1}{\sqrt{9}} \ln\left|\frac{{4x+1} - \sqrt{9}}{{4x+1} + \sqrt{{9}}}\right|+C \\
&= \frac{1}{{3}} \ln\left|\frac{{2x-1}}{{2x+2}}\right|+C
\end{align*} ∫ Q 1 d x = ∫ 2 x 2 + x − 1 1 d x = D 1 ln Q ′ + D Q ′ − D + C = 9 1 ln 4 x + 1 + 9 4 x + 1 − 9 + C = 3 1 ln 2 x + 2 2 x − 1 + C 2. Integration of p x + q a x 2 + b x + c \frac{px+q}{{a{x^2} + bx + c}} a x 2 + b x + c p x + q
First of all, let's assume the linear and the quadratic expression as L = p x + q L=px+q L = p x + q Q = a x 2 + b x + c Q = ax^2 + bx + c Q = a x 2 + b x + c bridge coefficient and will denote it with the symbol Ƶ Ƶ Ƶ
Ƶ = p b − 2 a q Ƶ=pb-2aq Ƶ = p b − 2 a q Based on these definitions, the formula I derived for the integral is as follows:
∫ p x + q a x 2 + b x + c = ∫ L Q d x \int \frac{px+q}{{a{x^2} + bx + c}} = \int \frac{L}{Q} dx ∫ a x 2 + b x + c p x + q = ∫ Q L d x ∫ L Q d x = p ln ∣ Q ∣ − Ƶ ∫ 1 Q d x 2 a + C \int \frac{L}{Q} dx = \frac{p\ln|Q| - Ƶ \int \frac{1}{Q} dx}{2a} + C ∫ Q L d x = 2 a p ln ∣ Q ∣ − Ƶ ∫ Q 1 d x + C Question 3. Evaluate the following integral:
∫ 3 x − 1 x 2 + 10 x + 28 d x \int{{\frac{{3x - 1}}{{{x^2} + 10x + 28}}dx}} ∫ x 2 + 10 x + 28 3 x − 1 d x Using step by step method: x x x
So, let’s again complete the square on the denominator and see what we get,
x 2 + 10 x + 28 = x 2 + 10 x + 25 − 25 + 28 = ( x + 5 ) 2 + 3 \begin{align*}
x^2 + 10x + 28 &= x^2 + 10x + 25 - 25 + 28 \\
&= \left( x + 5 \right)^2 + 3
\end{align*} x 2 + 10 x + 28 = x 2 + 10 x + 25 − 25 + 28 = ( x + 5 ) 2 + 3 Upon completing the square the integral becomes:
∫ 3 x − 1 x 2 + 10 x + 28 d x = ∫ 3 x − 1 ( x + 5 ) 2 + 3 d x \int \frac{3x - 1}{x^2 + 10x + 28} dx = \int \frac{3x - 1}{\left( x + 5 \right)^2 + 3} dx ∫ x 2 + 10 x + 28 3 x − 1 d x = ∫ ( x + 5 ) 2 + 3 3 x − 1 d x At this point we can use the same type of substitution that we did in the question 1. The only real difference is that we’ll need to make sure that we plug the substitution back into the numerator as well.
u = x + 5 x = u − 5 d x = d u u = x + 5 \quad x = u - 5 \quad dx = du u = x + 5 x = u − 5 d x = d u ∫ 3 x − 1 x 2 + 10 x + 28 d x = ∫ 3 ( u − 5 ) − 1 u 2 + 3 d u = ∫ 3 u u 2 + 3 − 16 u 2 + 3 d u = 3 2 ln ∣ u 2 + 3 ∣ − 16 3 tan − 1 ( u 3 ) + C = 3 2 ln ∣ ( x + 5 ) 2 + 3 ∣ − 16 3 tan − 1 ( x + 5 3 ) + C \begin{align*}
\int \frac{3x - 1}{x^2 + 10x + 28} dx & = \int \frac{3 \left( u - 5 \right) - 1}{u^2 + 3} du \\
& = \int \frac{3u}{u^2 + 3} - \frac{16}{u^2 + 3} du \\
& = \frac{3}{2} \ln \left| u^2 + 3 \right| - \frac{16}{\sqrt{3}} \tan^{-1} \left( \frac{u}{\sqrt{3}} \right) + C \\
& = \frac{3}{2} \ln \left| \left( x + 5 \right)^2 + 3 \right| - \frac{16}{\sqrt{3}} \tan^{-1} \left( \frac{x + 5}{\sqrt{3}} \right) + C
\end{align*} ∫ x 2 + 10 x + 28 3 x − 1 d x = ∫ u 2 + 3 3 ( u − 5 ) − 1 d u = ∫ u 2 + 3 3 u − u 2 + 3 16 d u = 2 3 ln u 2 + 3 − 3 16 tan − 1 ( 3 u ) + C = 2 3 ln ( x + 5 ) 2 + 3 − 3 16 tan − 1 ( 3 x + 5 ) + C Using the new formula: x 2 + 10 x + 28 x^2 + 10x + 28 x 2 + 10 x + 28 Q Q Q Q ′ = 2 x + 10 Q' = 2x+10 Q ′ = 2 x + 10 D = b 2 − 4 a c D=b^2-4ac D = b 2 − 4 a c
D = b 2 − 4 a c = ( 10 ) 2 − 4 ( 1 ) ( 28 ) = − 12 \begin{align*}
D &=b^2-4ac \\
&= (10)^2-4(1)(28) \\
&= -12
\end{align*} D = b 2 − 4 a c = ( 10 ) 2 − 4 ( 1 ) ( 28 ) = − 12 Now let's put the values of variables in the formula for D < 0 D<0 D < 0
∫ 1 Q d x = ∫ 1 x 2 + 10 x + 28 d x = 2 ∣ D ∣ tan − 1 ( Q ′ ∣ D ∣ ) + C = 2 12 tan − 1 ( 2 x + 10 12 ) + C = 1 3 tan − 1 ( x + 5 3 ) + C \begin{align*}
\int \frac{1}{Q} dx &= \int \frac{1}{x^2 + 10x + 28} dx \\
&= {\frac{2}{\sqrt{|D|}}{\tan^{-1}\left(\frac{Q'}{\sqrt{|D|}}\right)}} + C \\
&= \frac{2}{\sqrt{12}} \tan^{-1} \left( \frac{2x+10}{\sqrt{12}} \right) + C \\
&= \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{x+5}{\sqrt{3}} \right) + C
\end{align*} ∫ Q 1 d x = ∫ x 2 + 10 x + 28 1 d x = ∣ D ∣ 2 tan − 1 ( ∣ D ∣ Q ′ ) + C = 12 2 tan − 1 ( 12 2 x + 10 ) + C = 3 1 tan − 1 ( 3 x + 5 ) + C On comparing p x + q a x 2 + b x + c \frac{px+q}{{a{x^2} + bx + c}} a x 2 + b x + c p x + q 3 x − 1 x 2 + 10 x + 28 {{\frac{{3x - 1}}{{{x^2} + 10x + 28}}}} x 2 + 10 x + 28 3 x − 1 p = 3 p=3 p = 3 q = − 1 q=-1 q = − 1 a = 1 a=1 a = 1 b = 10 b=10 b = 10 Ƶ Ƶ Ƶ
Ƶ = p b − 2 a q = ( 3 ) ( 10 ) − 2 ( 1 ) ( − 1 ) = 32 \begin{align*}
Ƶ &=pb-2aq \\
&=(3)(10)-2(1)(-1) \\
&=32
\end{align*} Ƶ = p b − 2 a q = ( 3 ) ( 10 ) − 2 ( 1 ) ( − 1 ) = 32 So, let's plug all these values in the formula and see what we get :)
∫ L Q d x = ∫ 3 x − 1 x 2 + 10 x + 28 d x = p ln ∣ Q ∣ − Ƶ ∫ 1 Q d x 2 a + C = 3 ln ∣ x 2 + 10 x + 28 ∣ − 32 1 3 tan − 1 ( x + 5 3 ) + C 2 ( 1 ) + C = 3 2 ln ∣ x 2 + 10 x + 28 ∣ − 16 3 tan − 1 ( x + 5 3 ) + C \begin{align*}
\int \frac{L}{Q} dx &= \int \frac{3x - 1}{x^2 + 10x + 28} dx \\
&= \frac{p\ln|Q| - Ƶ \int \frac{1}{Q} dx}{2a} + C \\
&= \frac{3\ln|x^2 + 10x + 28| - 32 \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{x+5}{\sqrt{3}} \right)+C}{2(1)} + C \\
&= \frac{3}{2} \ln |x^2 + 10x + 28| - \frac{16}{\sqrt{3}} \tan^{-1} \left( \frac{x + 5}{\sqrt{3}} \right) + C
\end{align*} ∫ Q L d x = ∫ x 2 + 10 x + 28 3 x − 1 d x = 2 a p ln ∣ Q ∣ − Ƶ ∫ Q 1 d x + C = 2 ( 1 ) 3 ln ∣ x 2 + 10 x + 28∣ − 32 3 1 tan − 1 ( 3 x + 5 ) + C + C = 2 3 ln ∣ x 2 + 10 x + 28∣ − 3 16 tan − 1 ( 3 x + 5 ) + C Alright, let's solve another question using the formula where the discriminant of the quadratic equation will be positive so that we can get the answer of ∫ 1 Q d x \int \frac{1}{Q}dx ∫ Q 1 d x l n ln l n
Question 4. Evaluate the following integral:
∫ x + 3 x 2 − 2 x − 5 d x \int{{\frac{{x + 3}}{{{x^2} -2x-5}}dx}} ∫ x 2 − 2 x − 5 x + 3 d x Let's assume the quadratic x 2 − 2 x − 5 x^2 -2x-5 x 2 − 2 x − 5 Q Q Q
Q ′ = 2 x − 2 Q' = 2x-2 Q ′ = 2 x − 2 D = b 2 − 4 a c = ( − 2 ) 2 − 4 ( 1 ) ( − 5 ) = 24 \begin{align*}
D &=b^2-4ac \\
&= (-2)^2-4(1)(-5) \\
&= 24
\end{align*} D = b 2 − 4 a c = ( − 2 ) 2 − 4 ( 1 ) ( − 5 ) = 24 Now let's put the values of variables in the formula for D > 0 D>0 D > 0
∫ 1 Q d x = ∫ 1 x 2 − 2 x − 5 d x = 1 D ln ∣ Q ′ − D Q ′ + D ∣ + C = 1 24 ln ∣ 2 x − 2 − 24 2 x − 2 + 24 ∣ + C = 1 2 6 ln ∣ x − 1 − 6 x − 1 + 6 ∣ + C \begin{align*}
\int \frac{1}{Q} dx &= \int \frac{1}{x^2 -2x-5} dx \\
&= \frac{1}{\sqrt{D}} \ln\left|\frac{{Q'} - \sqrt{D}}{{Q'} + \sqrt{{D}}}\right|+C \\
&= \frac{1}{\sqrt{24}} \ln\left|\frac{{2x-2} - \sqrt{24}}{{2x-2} + \sqrt{{24}}}\right|+C \\
&= \frac{1}{2\sqrt{6}} \ln\left|\frac{{x-1} - \sqrt{6}}{{x-1} + \sqrt{{6}}}\right|+C
\end{align*} ∫ Q 1 d x = ∫ x 2 − 2 x − 5 1 d x = D 1 ln Q ′ + D Q ′ − D + C = 24 1 ln 2 x − 2 + 24 2 x − 2 − 24 + C = 2 6 1 ln x − 1 + 6 x − 1 − 6 + C On comparing p x + q a x 2 + b x + c \frac{px+q}{{a{x^2} + bx + c}} a x 2 + b x + c p x + q x + 3 x 2 − 2 x − 5 {{\frac{{x+3}}{{{x^2} -2x-5}}}} x 2 − 2 x − 5 x + 3 p = 1 p=1 p = 1 q = 3 q=3 q = 3 a = 1 a=1 a = 1 b = − 2 b=-2 b = − 2
Ƶ = p b − 2 a q = ( 1 ) ( − 2 ) − 2 ( 1 ) ( 3 ) = − 8 \begin{align*}
Ƶ &=pb-2aq \\
&=(1)(-2)-2(1)(3) \\
&=-8
\end{align*} Ƶ = p b − 2 a q = ( 1 ) ( − 2 ) − 2 ( 1 ) ( 3 ) = − 8 So, let's plug all these values in the formula and see what we get :)
∫ L Q d x = ∫ x + 3 x 2 − 2 x − 5 d x = p ln ∣ Q ∣ − Ƶ ∫ 1 Q d x 2 a + C = 1 ln ∣ x 2 − 2 x − 5 ∣ − ( − 8 ) 1 2 6 ln ∣ x − 1 − 6 x − 1 + 6 ∣ + C 2 ( 1 ) + C = 1 2 ln ∣ x 2 − 2 x − 5 ∣ + 2 6 ln ∣ x − 1 − 6 x − 1 + 6 ∣ + C \begin{align*}
\int \frac{L}{Q} dx &= \int \frac{x+3}{x^2 -2x-5} dx \\
&= \frac{p\ln|Q| - Ƶ \int \frac{1}{Q} dx}{2a} + C \\
&= \frac{1\ln|x^2 -2x-5| - (-8) \frac{1}{2\sqrt{6}} \ln\left|\frac{{x-1} - \sqrt{6}}{{x-1} + \sqrt{{6}}}\right|+C}{2(1)} + C \\
&= \frac{1}{2} \ln |x^2 -2x-5| + \frac{2}{\sqrt{6}} \ln\left|\frac{{x-1} - \sqrt{6}}{{x-1} + \sqrt{{6}}}\right|+C
\end{align*} ∫ Q L d x = ∫ x 2 − 2 x − 5 x + 3 d x = 2 a p ln ∣ Q ∣ − Ƶ ∫ Q 1 d x + C = 2 ( 1 ) 1 ln ∣ x 2 − 2 x − 5∣ − ( − 8 ) 2 6 1 ln x − 1 + 6 x − 1 − 6 + C + C = 2 1 ln ∣ x 2 − 2 x − 5∣ + 6 2 ln x − 1 + 6 x − 1 − 6 + C 3. Integration of 1 a x 2 + b x + c \frac{1}{\sqrt{a{x^2} + bx + c}} a x 2 + b x + c 1
First of all, let's assume the quadratic expression as Q = a x 2 + b x + c Q = ax^2 + bx + c Q = a x 2 + b x + c Q Q Q Q ′ = 2 a x + b Q' = 2ax + b Q ′ = 2 a x + b D D D D = b 2 − 4 a c D = b^2 - 4ac D = b 2 − 4 a c
∫ 1 a x 2 + b x + c d x = ∫ 1 Q d x \int \frac{1}{\sqrt{a{x^2} + bx + c}} dx = \int \frac{1}{\sqrt{Q}}dx ∫ a x 2 + b x + c 1 d x = ∫ Q 1 d x When ( a > 0 ) : (a>0): ( a > 0 ) : —
∫ 1 Q d x = 1 a ln ∣ 2 a Q + Q ′ 2 a ∣ + C \int \frac{1}{\sqrt{Q}}dx = \frac{1}{\sqrt{a}}{\ln\left| \frac{2 \sqrt{a} \sqrt{Q} + Q'}{2a}\right|} +C ∫ Q 1 d x = a 1 ln 2 a 2 a Q + Q ′ + C When ( a < 0 ) : (a<0): ( a < 0 ) : —
∫ 1 Q d x = 1 ∣ a ∣ sin − 1 ( Q ′ ∣ D ∣ ) + C \int \frac{1}{\sqrt{Q}}dx = {\frac{1}{\sqrt{|a|}}{\sin^{-1}\left(\frac{Q'}{\sqrt{|D|}}\right)}} + C ∫ Q 1 d x = ∣ a ∣ 1 sin − 1 ( ∣ D ∣ Q ′ ) + C incomplete...
