How do you find int (x^2+2x)/((x^2+1)^2)dx using partial fractions?

Feb 2, 2017

$\int \frac{2 x - 1}{{x}^{2} + 1} ^ 2 \mathrm{dx} = \frac{1}{2} \left({\tan}^{- 1} x\right) + \frac{x - 2}{{x}^{2} + 1} + c$

Explanation:

Partial fractions of $\frac{{x}^{2} + 2 x}{{x}^{2} + 1} ^ 2$, will be of type

$\frac{{x}^{2} + 2 x}{{x}^{2} + 1} ^ 2 = \frac{A x + B}{{x}^{2} + 1} + \frac{C x + D}{{x}^{2} + 1} ^ 2$

= $\frac{\left(A x + B\right) \left({x}^{2} + 1\right) + C x + D}{{x}^{2} + 1} ^ 2$

= $\frac{A {x}^{3} + B {x}^{2} + A x + B + C x + D}{{x}^{2} + 1} ^ 2$

= $\frac{A {x}^{3} + B {x}^{2} + \left(A + C\right) x + \left(B + D\right)}{{x}^{2} + 1} ^ 2$

Comparing the terms, we get $A = 0$, $B = 1$, $A + C = 2$ and $B + D = 0$

i.e. $A = 0$, $B = 1$, $C = 2$ and $D = - 1$

Hence, $\frac{{x}^{2} + 2 x}{{x}^{2} + 1} ^ 2 = \frac{1}{{x}^{2} + 1} + \frac{2 x - 1}{{x}^{2} + 1} ^ 2$ and

$\int \frac{{x}^{2} + 2 x}{{x}^{2} + 1} ^ 2 \mathrm{dx} = \int \frac{1}{{x}^{2} + 1} \mathrm{dx} + \int \frac{2 x - 1}{{x}^{2} + 1} ^ 2 \mathrm{dx}$

Now $\int \frac{1}{{x}^{2} + 1} \mathrm{dx} = {\tan}^{- 1} x$ is a well known integral and for the other one

We can break $\int \frac{2 x - 1}{{x}^{2} + 1} ^ 2 \mathrm{dx} = \int \frac{2 x}{{x}^{2} + 1} ^ 2 \mathrm{dx} - \int \frac{1}{{x}^{2} + 1} ^ 2 \mathrm{dx}$

For first term, let us assume $u = {x}^{2} + 1$, then $\mathrm{du} = 2 x \mathrm{dx}$= $\int \frac{2 x}{{x}^{2} + 1} ^ 2 \mathrm{dx} = \int \frac{\mathrm{du}}{u} ^ 2 = - \frac{1}{u} = - \frac{1}{{x}^{2} + 1}$

and for second term let us assume $x = \tan v$ then

$\int \frac{1}{{x}^{2} + 1} ^ 2 \mathrm{dx} = \int \frac{{\sec}^{2} v}{\sec} ^ 4 v \mathrm{dv} = \int {\cos}^{2} v \mathrm{dv}$

= $\frac{1}{2} \int \left(1 + \cos 2 v\right) \mathrm{dv} = \frac{v}{2} + \frac{1}{4} \sin 2 v$

= ${\tan}^{- 1} \frac{x}{2} + \frac{1}{2} \sin v \cos v$

= ${\tan}^{- 1} \frac{x}{2} + \frac{1}{2} \frac{x}{\sqrt{{x}^{2} + 1}} \times \frac{1}{\sqrt{{x}^{2} + 1}}$

= ${\tan}^{- 1} \frac{x}{2} + \frac{1}{2} \frac{x}{{x}^{2} + 1}$

Hence, $\int \frac{2 x - 1}{{x}^{2} + 1} ^ 2 \mathrm{dx}$

= $- \frac{1}{{x}^{2} + 1} + {\tan}^{- 1} \frac{x}{2} + \frac{1}{2} \frac{x}{{x}^{2} + 1} + c$

= $\frac{1}{2} \left({\tan}^{- 1} x\right) + \frac{x - 2}{{x}^{2} + 1} + c$