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

Oct 30, 2015

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

#### Explanation:

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

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

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

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

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

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

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

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

$= \frac{{x}^{4} \left(A + B + C + D\right) + {x}^{3} \left(B - C + E\right)}{x \left(x - 1\right) \left(x + 1\right) \left({x}^{2} + 1\right)} +$

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

$A + B + C + D = 0$
$B - C + E = 0$
$B + C - D = 0$
$B - C - E = 0$
$- A = 2 \implies A = - 2$

$\left[I I - I V\right] \implies 2 E = 0 \implies E = 0$

$B + C + D = 2$
$B - C = 0 \implies B = C$
$B + C - D = 0$

$2 C + D = 2$
$2 C - D = 0$

$4 C = 2 \implies C = \frac{1}{2} \implies B = \frac{1}{2}$

$D = 2 C \implies D = 1$

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

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

Note:

$\int \frac{x \mathrm{dx}}{{x}^{2} + 1} = \int \frac{\frac{1}{2} \left(2 x \mathrm{dx}\right)}{{x}^{2} + 1} = \frac{1}{2} \int \frac{d \left({x}^{2} + 1\right)}{{x}^{2} + 1} = \frac{1}{2} \ln | {x}^{2} + 1 |$