Question

Integrate the rational functions.
$\int \frac{1}{x(x^n+1)}dx$

Answer 1

Let $I=\int \frac{1}{x(x^n+1)}dx=\int \frac{x^{n-1}}{x^n(x^n+1)}dx$
Put $x^n=t\Rightarrow n{x}^{n-1}dx=dt\Rightarrow x^{n-1}dx=\frac{1}{n}dt$
$\therefore I=\int \frac{x^{n-1}}{x^n(x^n+1)}dx=\frac{1}{n}\int \frac{1}{t(t+1)}dt......\left(i\right)$
Now, $\frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{(t+1)}\Rightarrow 1=A(1+t)+Bt.....\left(ii\right)$
On substituting t =0, -1 in Eq. (ii), we get A =1 and B =-1
$\therefore \frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{(t+1)}$
$\therefore I=\frac{1}{n}\int (\frac{1}{t}-\frac{1}{(t+1)})dt$ [from Eq.(i)]
$=\frac{1}{n}[log|t|-log|t+1|]+C=\frac{1}{n}[log|x^n|-log|x^n+1|]+C\text{[put t=}{x}^n\text{]}=\frac{1}{n}log\left|\frac{x^n}{x^n+1}\right|+C$