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

Dec 19, 2016

$\frac{1}{4} \ln | x - 1 | + \frac{3}{4} \ln | x + 1 | + \frac{1}{2 \left(x + 1\right)} + C$

#### Explanation:

Let ${x}^{2} / \left(\left(x - 1\right) \left({x}^{2} + 2 x + 1\right)\right) \equiv \frac{\frac{1}{4}}{x - 1} + \frac{B x + C}{x + 1} ^ 2$ for all $x$.
(The $\frac{1}{4}$ comes from the cover-up rule: ${1}^{2} / \left({1}^{2} + 2 \times 1 + 1\right)$.)
Then, multiplying out,
$4 {x}^{2} \equiv {x}^{2} + 2 x + 1 + 4 B {x}^{2} + 4 \left(C - B\right) x - 4 C$
Collecting powers of $x$:
$4 {x}^{2} \equiv \left(1 + 4 B\right) {x}^{2} + \left(2 + 4 C + 4 B\right) x + \left(1 - 4 C\right)$
Equating powers of ${x}^{2}$ on both sides of the identity:
$4 = 1 + 4 B$ hence $B = \frac{3}{4}$.
Equating constants on both sides gives $C = \frac{1}{4}$.
Hence the integral is:
$\left(\frac{1}{4}\right) \int \left(\frac{1}{x - 1} + \frac{3 x + 1}{x + 1} ^ 2\right) \mathrm{dx}$
$= \frac{1}{4} \int \left(\frac{1}{x - 1} + \frac{3 x + 3 - 2}{x + 1} ^ 2\right) \mathrm{dx}$
$= \frac{1}{4} \int \left(\frac{1}{x - 1} + \frac{3 x + 3}{x + 1} ^ 2 - \frac{2}{x + 1} ^ 2\right) \mathrm{dx}$
$= \frac{1}{4} \ln | x - 1 | + \frac{3}{8} \ln {\left(x + 1\right)}^{2} + \frac{1}{2 \left(x + 1\right)} + C$