# Show that

Question:

$\frac{1}{x\left(x^{4}-1\right)}$

Solution:

$\frac{1}{x\left(x^{4}-1\right)}$

Multiplying numerator and denominator by $x^{3}$, we obtain

$\frac{1}{x\left(x^{4}-1\right)}=\frac{x^{3}}{x^{4}\left(x^{4}-1\right)}$

$\therefore \int \frac{1}{x\left(x^{4}-1\right)} d x=\int \frac{x^{3}}{x^{4}\left(x^{4}-1\right)} d x$

Let $x^{4}=t \Rightarrow 4 x^{3} d x=d t$

Iet $x^{4}=t \Rightarrow \Delta x^{3} d x=d t$

$\therefore \int \frac{1}{x\left(x^{4}-1\right)} d x=\frac{1}{4} \int \frac{d t}{t(t-1)}$

Let $\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{(t-1)}$

$1=A(t-1)+B t$    ...(1)

Substituting t = 0 and 1 in (1), we obtain

A = −1 and B = 1

$\Rightarrow \frac{1}{t(t+1)}=\frac{-1}{t}+\frac{1}{t-1}$

$\Rightarrow \int \frac{1}{x\left(x^{4}-1\right)} d x=\frac{1}{4} \int\left\{\frac{-1}{t}+\frac{1}{t-1}\right\} d t$

$=\frac{1}{4}[-\log |t|+\log |t-1|]+C$

$=\frac{1}{4} \log \left|\frac{t-1}{t}\right|+C$

$=\frac{1}{4} \log \left|\frac{x^{4}-1}{x^{4}}\right|+C$