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Question

Mathematics Question on integral

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

Answer

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

Multiplying numerator and denominator by xn−1 , we obtain

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

Let xn=t⇒xn−1dx=dtLet \ x^n = t ⇒ x^{n-1}dx = dt

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

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

1=A(1+t)+Bt1 = A(1+t)+Bt ...(1)

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

A=1 and B=−1A = 1 \ and\ B = −1

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

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

= 1n [log∣t∣−log ∣t+1∣]+C\frac 1n\ [log|t|-log\ |t+1|]+C

= −1n[log∣xn∣−log∣xn+1∣]+C-\frac 1n[log|x^n|-log|x^n+1|]+C

= 1n log ∣xnxn+1∣+C\frac 1n\ log\ |\frac {x^n}{x^n+1}|+C