# How do you integrate int (2x-2)/((x^4-1)(x-3))dx using partial fractions?

Mar 2, 2017

$\int \frac{2 x - 2}{\left({x}^{4} - 1\right) \left(x - 3\right)} \mathrm{dx} = - \frac{1}{4} \ln | x + 1 | + \frac{1}{10} \ln \left({x}^{2} + 1\right) - \frac{2}{5} {\tan}^{- 1} x + \frac{1}{20} \ln | x - 3 |$

#### Explanation:

Let us first obtain partial fractions of $\frac{2 x - 2}{\left({x}^{4} - 1\right) \left(x - 3\right)}$ and as ${x}^{4} - 1 = \left(x + 1\right) \left(x - 1\right) \left({x}^{2} + 1\right)$, these will be

$\frac{2 x - 2}{\left({x}^{4} - 1\right) \left(x - 3\right)} = \frac{2 \left(x - 1\right)}{\left(x + 1\right) \left(x - 1\right) \left({x}^{2} + 1\right) \left(x - 3\right)} = \frac{2}{\left(x + 1\right) \left({x}^{2} + 1\right) \left(x - 3\right)}$

and this can be expressed as

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

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

putting $x = - 1$ and $x = 3$ in this we get

$2 = - 8 A$ i.e. $A = - \frac{1}{4}$ and $40 D = 2$ i.e. $D = \frac{1}{20}$

and comparing coefficients of ${x}^{3}$ and constant term,

we have, $0 = A + B + D$ i.e. $B = - A - D = \frac{1}{4} - \frac{1}{20} = \frac{4}{20} = \frac{1}{5}$

and $2 = - 3 A - 3 C + D$ or $3 C = - 2 - 3 \left(- \frac{1}{4}\right) + \frac{1}{20} = - 2 + \frac{3}{4} + \frac{1}{20} = \frac{- 24}{20} = - \frac{6}{5}$ and $C = - \frac{2}{5}$ and

$\frac{2}{\left(x + 1\right) \left({x}^{2} + 1\right) \left(x - 3\right)} = - \frac{1}{4 \left(x + 1\right)} + \frac{x - 2}{5 \left({x}^{2} + 1\right)} + \frac{1}{20 \left(x - 3\right)}$

and $\int \frac{2 x - 2}{\left({x}^{4} - 1\right) \left(x - 3\right)} \mathrm{dx}$

= $\int - \frac{1}{4 \left(x + 1\right)} \mathrm{dx} + \int \frac{x - 2}{5 \left({x}^{2} + 1\right)} \mathrm{dx} + \int \frac{1}{20 \left(x - 3\right)} \mathrm{dx}$

= $\int - \frac{1}{4 \left(x + 1\right)} \mathrm{dx} + \frac{1}{5} \left(\int \frac{x}{{x}^{2} + 1} \mathrm{dx} - \int \frac{2}{{x}^{2} + 1} \mathrm{dx}\right) + \int \frac{1}{20 \left(x - 3\right)} \mathrm{dx}$

= $- \frac{1}{4} \ln | x + 1 | + \frac{1}{5} \left(\frac{1}{2} \ln \left({x}^{2} + 1\right) - 2 {\tan}^{- 1} x\right) + \frac{1}{20} \ln | x - 3 |$

= $- \frac{1}{4} \ln | x + 1 | + \frac{1}{10} \ln \left({x}^{2} + 1\right) - \frac{2}{5} {\tan}^{- 1} x + \frac{1}{20} \ln | x - 3 |$