Question

Integrate the rational function: \(\frac {1}{x(x^n+1)}\)

Hint:

Multiply numerator and denominator by xⁿ⁻¹ and put xⁿ = t

Answer 1

\(\frac {1}{x(x^n+1)}\)

Multiplying numerator and denominator by xⁿ⁻¹ , we obtain

\(\frac {1}{x(x^n+1)}\) = \(\frac {x^{n-1}}{x^{n-1}x(x^n+1)}\) = \(\frac {x^{n-1}}{x^n(x^n+1)}\)

\(Let \ x^n = t ⇒ x^{n-1}dx = dt\)

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

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

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

\(Substituting\  t = 0,−1 \ in\  equation\  (1), we\  obtain\)

\(A = 1 \ and\  B = −1\)

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

⇒ \(∫\)\(\frac {1}{x(x^n+1)}\ dx\) = \(\frac 1n\) \(∫\)\([\frac 1t-\frac {1}{(1+t)} ]\ dx\)

= \(\frac 1n\ [log|t|-log\ |t+1|]+C\)

= \(-\frac 1n[log|x^n|-log|x^n+1|]+C\)

= \(\frac 1n\  log\ |\frac {x^n}{x^n+1}|+C\)