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

May 2, 2017

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

#### Explanation:

Partial fractions of $\frac{x + 2}{\left({x}^{2} + 2\right) x}$ would be of the form

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

I.e. $x + 2 = A \left({x}^{2} + 2\right) + x \left(B x + C\right)$

Comparing coefficients of $x \cdot 2$, $x$ and constant term

we have $A + B = 0$, $C = 1$ and $2 A = 2$

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

i.e. $\frac{x + 2}{\left({x}^{2} + 2\right) x} = \frac{1}{x} - \frac{x - 1}{{x}^{2} + 2}$

Hence our integral becomes $\int \left(\frac{1}{x} - \frac{x - 1}{{x}^{2} + 2}\right) \mathrm{dx}$

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

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