# How do you evaluate the integral int 1/(x(x-1)^3)?

Mar 14, 2018

$\int \frac{1}{x {\left(x - 1\right)}^{3}} \mathrm{dx} = - \ln \left\mid x \right\mid + \ln \left\mid x - 1 \right\mid + \frac{1}{x - 1} - \frac{1}{2 {\left(x - 1\right)}^{2}} + C$

#### Explanation:

Write:

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

Multiplying both sides by $x {\left(x - 1\right)}^{3}$ we get:

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

Putting $x = 0$ we find $A = - 1$

Looking at the coefficient of ${x}^{3}$, we find $B = - A = 1$

Putting $x = 1$ we find $D = 1$

Looking at the coefficient of $x$, we find:

$0 = 3 A + B - C + D$

Hence $C = - 1$

So

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

and:

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

$\textcolor{w h i t e}{\int \frac{1}{x {\left(x - 1\right)}^{3}} \mathrm{dx}} = - \ln \left\mid x \right\mid + \ln \left\mid x - 1 \right\mid + \frac{1}{x - 1} - \frac{1}{2 {\left(x - 1\right)}^{2}} + C$