# How do you solve this integral?

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

Mar 28, 2018

$\int \setminus \frac{\text{d} x}{{x}^{2} - 1} ^ 2$
$= \frac{1}{4} \left(\ln \left(x + 1\right) - \ln \left(x - 1\right) - \frac{2 x}{{x}^{2} - 1}\right) + C$

#### Explanation:

$\int \setminus \frac{\text{d} x}{{x}^{2} - 1} ^ 2$
$= \int \setminus \frac{\text{d} x}{{\left(x + 1\right)}^{2} {\left(x - 1\right)}^{2}}$

Now, let's do the partial fractions. Assume that

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

for some constants $A , B , C , D$.

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

Expand to get
$1 = \left(A + C\right) {x}^{3} + \left(B + C + D - A\right) {x}^{2} + \left(2 D - 2 B - A - C\right) x + A + B - C + D$.

Equate coefficients:
$\left\{\begin{matrix}A + C = 0 \\ B + C + D - A = 0 \\ 2 D - 2 B - A - C = 0 \\ A + B - C + D = 1\end{matrix}\right.$

Solving gives $A = B = D = \frac{1}{4}$ and $C = - \frac{1}{4}$.

Thus, our original integral is
$\int \setminus \left(\frac{1}{4 \left(x + 1\right)} + \frac{1}{4 {\left(x + 1\right)}^{2}} - \frac{1}{4 \left(x - 1\right)} + \frac{1}{4 {\left(x - 1\right)}^{2}}\right) \setminus \text{d} x$
$= \frac{1}{4} \ln \left(x + 1\right) - \frac{1}{4 \left(x + 1\right)} - \frac{1}{4} \ln \left(x - 1\right) - \frac{1}{4 \left(x - 1\right)} + C$

Simplify:
$= \frac{1}{4} \left(\ln \left(x + 1\right) - \ln \left(x - 1\right) - \frac{2 x}{{x}^{2} - 1}\right) + C$