Question

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

Answer 1

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

Multiplying numerator and denominator by x³ , we obtain

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

∴ \(∫\)\(\frac {1}{x(x^4-1)}dx\) = \(∫\)\(\frac {x^3}{x^4(x^4-1)}dx\)

Let x⁴ = t ⇒ 4x³dx = dt

∴ \(∫\)\(\frac {1}{x(x^4-1)}dx\) = \(\frac 14∫\frac {dt}{t(t-1)}\)

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

\(1 = A(t-1) + Bt   \)                       ...(1)

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

\(A = -1\  and \ B = 1\)

⇒ \(\frac {1}{t(t-1)}\) = \(\frac {-1}{t}+\frac {1}{t-1}\)

⇒ \(∫\)\(\frac {1}{x(x^4-1)}dx\) = \(\frac 14\) \(∫\)\(\frac {-1}{t}+\frac {1}{t-1}dt\)

= \(\frac 14[-log|t|+log|t-1|]+C\)

= \(\frac 14log\ |\frac {t-1}{t}|+C\)

= \(\frac 14log|\frac {x^4-1}{x^4}|+C\)