# How do you use partial fractions to find the integral int (x^2-1)/(x^3+x)dx?

Feb 28, 2017

$\ln | \frac{{x}^{2} + 1}{x} | + C$

#### Explanation:

Start by factoring the denominator.

${x}^{3} + x = x \left({x}^{2} + 1\right)$

Now write the partial fraction decomposition.

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

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

$A {x}^{2} + B x + C {x}^{2} + C = {x}^{2} - 1$

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

We now write a system of equations.

$\left\{\begin{matrix}A + C = 1 \\ B = 0 \\ C = - 1\end{matrix}\right.$

Solving, we get $A = 2 , B = 0 , C = - 1$.

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

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

We now make the substitution $u = {x}^{2} + 1$. Then $\mathrm{du} = 2 x \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{du}}{2 x}$.

$\int \frac{2 x}{u} \cdot \frac{\mathrm{du}}{2 x} - \int \frac{1}{x} \mathrm{dx}$

$\int \frac{1}{u} \mathrm{du} - \int \frac{1}{x} \mathrm{dx}$

$\ln | u | - \ln | x | + C$

$\ln | {x}^{2} + 1 | - \ln | x | + C$

$\ln | \frac{{x}^{2} + 1}{x} | + C$

Hopefully this helps!