Question

Integrate the rational function $\frac{1}{x(x^4-1)}$

Answer 1

$\frac{1}{x(x^4-1)}$
Multiplying numerator and denominator by $x^3$, we obtain
$\frac{1}{x(x^4-1)}=\frac{x^3}{x^4(x^4-1)}$
$\therefore \int \frac{1}{x(x^4-1)}dx=\int \frac{x^3}{x^4(x^4-1)}dx$
Let $x^4=t\Rightarrow 4x^{3}dx=dt$
$\therefore \int \frac{x^3}{x^4(x^4-1)}dx=\frac{1}{4}\int \frac{dt}{t(t-1)}$
Let $\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{(t-1)}$
$\Rightarrow 1=A(t-1)+Bt$ ........... (1)
Substituting $=0$ and 1 in (1), we obtain
$A=-1$ and $B=1$
$\Rightarrow \frac{1}{t(t-1)}=\frac{-1}{t}+\frac{t}{t-1}$
$\Rightarrow \int \frac{1}{x(x^4-1)}dx=\frac{1}{4}\int \{\frac{-1}{t}+\frac{1}{t-1}\}dt$
$=\frac{1}{4}[-\log |t|+\log |t-1|]+C$
$=\frac{1}{4}\log \left|\frac{t-1}{t}\right|+C$
$=\frac{1}{4}\log \left|\frac{x^4-1}{x^4}\right|+C$