Home » Blog » Evaluate the integral of 1 / (x2 – 1)2

Evaluate the integral of 1 / (x2 – 1)2

by RoRi

Compute the following integral.

    \[ \int \frac{dx}{(x^2 -1)^2}. \]

First, we write

    \[ \int \frac{dx}{(x^2-1)^2} = \int \frac{dx}{(x+1)^2(x-1)^2}. \]

Then we use partial fractions, writing

    \[ \frac{1}{(x+1)^2(x-1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-1} + \frac{D}{(x-1)^2}. \]

This gives us the equation

    \[ A(x+1)(x-1)^2 + B(x-1)^2 + C(x+1)^2(x-1) + D(x+1)^2 = 1. \]

Evaluating at x = -1 and x = 1 we obtain the values for B and D,

    \begin{align*} 4B &= 1 & \implies \qquad B &= \frac{1}{4} \\ 4D &= 1 & \implies \qquad D &= \frac{1}{4}. \end{align*}

Using these values of B and D and evaluating at x = 0 and x = 2 we have

    \begin{align*} A + \frac{1}{4} - C + \frac{1}{4} &= 1 \\ 3A + \frac{1}{4} + 9C + \frac{9}{4} &= 1. \end{align*}

Solving this system for A and C we get

    \[ A = \frac{1}{4}, \qquad C = -\frac{1}{4}. \]

Thus,

    \begin{align*} \int \frac{dx}{(x^2-1)^2} &= \int \left( \frac{A}{(x+1)} + \frac{B}{(x+1)^2} + \frac{C}{x-1} + \frac{D}{(x-1)^2} \right) \, dx \\[10pt] &= \frac{1}{4} \left(\int \frac{1}{x+1} \, dx + \int \frac{1}{(x+1)^2} \, dx - \int \frac{1}{x-1} \,dx + \int \frac{1}{(x-1)^2} \, dx \\[10pt] &= \frac{1}{4} \left( \log|x+1| -\frac{1}{x+1} - \log|x-1| - \frac{1}{x-1}\right) + C \\[10pt] &= \frac{1}{4} \log \left| \frac{x+1}{x-1} \right| - \frac{1}{4} \left( \frac{1}{x+1} + \frac{1}{x-1} \right) + C \\[10pt] &= \frac{1}{4} \log \left| \frac{x+1}{x-1} \right| - \frac{x}{2(x^2-1)} + C. \end{align*}

One comment

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):