Question

Integrate the rational function:

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

Answer 1

Let ${\displaystyle I=\int \frac{1}{x(x^4-1)} dx}$

${\displaystyle =\frac{1}{4}\int \frac{4x^3}{x^4(x^4-1)} dx}$

Put x⁴ = t

⇒ 4x³ dx = dt

∴ ${\displaystyle I=\frac{1}{4}\int \frac{dt}{t(t-1)}}$

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

⇒ 1 = A (t - 1) + Bt               ....(i)

Putting t = 0 in (i), we get

1 = A (-1)

⇒ A = -1

Putting t = 1 in (i), we get

1 = B (1)

⇒ B = 1

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

∴ ${\displaystyle I=\frac{1}{4}\int (-\frac{1}{t}+\frac{1}{t-1}) dt}$

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

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

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