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

Oct 18, 2016

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

#### Explanation:

$\frac{{x}^{2} - 1}{x \left({x}^{2} + 1\right)} = \frac{A}{x} + \frac{B x + C}{{x}^{2} + 1}$
Developing
(x^2-1)/(x(x^2+1))=(A(x^2+1)+ x(Bx+C))/(x(x^2+1)
So we can compare the coefficients
${x}^{2} - 1 = A \left({x}^{2} + 1\right) + x \left(B x + C\right)$
For ${x}^{2}$ , $1 = A + B$ and $- 1 = A$ and $0 = C$
Hence $A = - 1$and $B = 2$
$\int \frac{\left({x}^{2} - 1\right) \mathrm{dx}}{x \left({x}^{2} + 1\right)} = \int A \frac{\mathrm{dx}}{x} + \int \frac{\left(B x + C\right) \mathrm{dx}}{{x}^{2} + 1}$
=int(-1dx)/x+int((2x)dx)/(x^2+1
$= - \ln x + \ln \left({x}^{2} + 1\right) + C$
$= \ln \left(\frac{{x}^{2} + 1}{x}\right) + C$