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

1 Answer
May 22, 2016

By converting the fraction into a sum of independent factors

Explanation:

We know how to integrate functions of the form $\frac{1}{x} ^ n$ or $\frac{1}{x - a} ^ n$ so we want to transform the complicated fraction into a sum of simple fractions. We write
$I = \frac{A}{x - 1} + \frac{B}{x - 1} ^ 2 + \frac{C}{{x}^{2} + 1}$
The constant term in the numerator is $B + C - A$ and is equal to $- 1$
The ${x}^{4}$ term is $B + C + A$ and must be zero. This gives us
$- 2 A = - 1$ or $A = - \frac{1}{2}$
The ${x}^{3}$ term is $- 2 C + A + B = 0$

Next $- 2 C + A + B + 2 C = 0 + 2 \left(- A - B\right)$ so
$B = - A = \frac{1}{2}$ and thus $C = 0$

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

Integral $= - \frac{1}{2} \ln \left(x - 1\right) - \frac{1}{2} {\left(x - 1\right)}^{2}$