# How do you use partial fractions to find the integral int (x^2+x+2)/(x^2+2)^2dx?

Mar 4, 2018

The answer is $= \frac{1}{\sqrt{2}} \arctan \left(\frac{x}{\sqrt{2}}\right) - \frac{1}{2 \left({x}^{2} + 2\right)} + C$

#### Explanation:

We need

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

Perform the decomposition into partial fractions

$\frac{{x}^{2} + x + 2}{{x}^{2} + 2} ^ 2 = \frac{{x}^{2} + 2 + x}{{x}^{2} + 2} ^ 2$

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

Therefore,

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

The first integral is

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

$= \frac{1}{2} \cdot \sqrt{2} \arctan \left(\frac{x}{\sqrt{2}}\right)$

$= \frac{1}{\sqrt{2}} \arctan \left(\frac{x}{\sqrt{2}}\right)$

For the second integral

Let $u = {x}^{2} + 2$, $\implies$, $\mathrm{du} = 2 x \mathrm{dx}$

Therefore,

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

$= - \frac{1}{2} \cdot \frac{1}{u}$

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

Finally,

$\int \frac{\left({x}^{2} + x + 2\right) \mathrm{dx}}{{x}^{2} + 2} ^ 2 = \frac{1}{\sqrt{2}} \arctan \left(\frac{x}{\sqrt{2}}\right) - \frac{1}{2 \left({x}^{2} + 2\right)} + C$