Let's Integrate!

$\dfrac{1}{x(x^n+1)}=\dfrac{x^{-n-1}}{1+x^{-n}}$ , and let $f(x)=1+x^{-n}$ , then $\dfrac{x^{-n-1}}{1+x^{-n}}=-\dfrac{f'(x)}{nf(x)}$ because $f'(x)=-nx^{-n-1}$ . Since

• $\displaystyle\int \dfrac{f'(x)}{f(x)}dx=\ln|f(x)|+C$ ,

We get the answer.

Using partial fraction decomposition,

$\displaystyle \frac{1}{x(x^n+1)} = \frac{A}{x}+\frac{Bx^{n-1}}{x^n+1}$

$\displaystyle 1=A(x^n+1)+Bx(x^{n-1})=Ax^n+Bx^n+A$

$A$ must be equal to $1$ and $Ax^n+Bx^n$ must be equal to $0$ , so

$x^n + Bx^n =0\,\,\,\Rightarrow\,\,\,B=-1$

Then our new integral is

$\displaystyle\int \left(\frac{1}{x}-\frac{x^{n-1}}{x^n+1}\right)\,\mathrm{d}x$

$u$ -substituting for the second term,

$\displaystyle u=x^n+1\,\,\,\Rightarrow\,\,\, \frac{1}{n}\mathrm{d}u=x^{n-1}\,\mathrm{d}x$

$\displaystyle \Rightarrow\,\int \frac{1}{x}\mathrm{d}x -\frac{1}{n}\int\frac{1}{u}\mathrm{d}u$

$\displaystyle = \ln x - \frac{1}{n}\ln(x^n+1) +C = \frac{1}{n}\ln x^n - \frac{1}{n}\ln(x^n+1) +C$

$\displaystyle = \displaystyle \boxed{\displaystyle \frac{1}{n}\ln \left( \frac{x^n}{x^n+1} \right) +C}$

write x^n + 1 - x^n on the numerator ... then separate and integrate...

put k = x*(n)......... then go by partial fraction method