Question

Integrate the rational functions.
$\int \frac{1}{x(x^4-1)}dx.$

Answer 1

Let $I=\int \frac{1}{x(x^4-1)}dx=\int \frac{x^3}{x^4(x^4-1)}dx=\int \frac{x^3}{x^4(x^4-1)}dx=\frac{1}{4}\int \frac{4x^3}{x^4(x^4-1)}dx$
Put $x^4=t\Rightarrow 4x^3=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{4x^3}$
$\therefore I=\frac{1}{4}\int \frac{4x^3}{t(t-1)}\frac{dt}{4x^3}=\frac{1}{4}\int \frac{dt}{t(t-1)}$
Let $\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{(t-1)}\Rightarrow 1=A(t-1)+Bt.....\left(i\right)$
On substituting t =0 and 1 in Eq. (i), we get
A=-1 and B =1
$\therefore \frac{1}{t(t-1)}=\frac{-1}{1}+\frac{1}{t-1}\therefore I=\frac{1}{4}\int (-\frac{1}{t}+\frac{1}{t-1})dt=\frac{1}{4}-log|t||t-1|+C=\frac{1}{4}log\left|\frac{t-1}{t}\right|+C=\frac{1}{4}log\left|\frac{x^4-1}{x^4}\right|+C\text{(Put t=}{x}^4\text{)}$