# How do you integrate 7/(x^2+1) using partial fractions?

Aug 21, 2016

$\int \frac{7}{{x}^{2} + 1} \mathrm{dx} = \frac{7}{2} \ln \left(x + i\right) - \frac{7}{2} \ln \left(x - i\right) + C$

#### Explanation:

This integral would normally be done using trigonometric substitution, but if you really want to use partial fractions to integrate this then you will need to use Complex coefficients:

$\frac{7}{{x}^{2} + 1} = \frac{A}{x + i} + \frac{B}{x - i} = \frac{A \left(x - i\right) + B \left(x + i\right)}{{x}^{2} + 1}$

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

Equating coefficients:

$\left\{\begin{matrix}A + B = 0 \\ \left(B - A\right) i = 7\end{matrix}\right.$

Multiply both sides of the second equation by $i$ to get:

$A - B = 7 i$

Hence $A = \frac{7}{2} i$ and $B = - \frac{7}{2} i$
So:

$\int \frac{7}{{x}^{2} + 1} \mathrm{dx} = \int \frac{7}{2 \left(x + i\right)} - \frac{7}{2 \left(x - i\right)} \mathrm{dx}$

$= \frac{7}{2} \ln \left(x + i\right) - \frac{7}{2} \ln \left(x - i\right) + C$