Question: $$\int \frac{dx}{x(x^n +1)} $$
$$t=x^n + 1$$
$${dt}=nx^{n-1}{dx}$$
$$\frac{dx}{x}=\frac{dt}{nx^n}$$
$$\int \frac{1}{n}\frac{dt}{t(t-1)} $$
I tried $t=\sin^2\theta $ but back substitution will be difficult .

How to integrate this type of integration ?

Answer: 
Let $$I = \int\frac{1}{x(x^n+1)}dx = \int\frac{1}{\left(1+x^{-n}\right)}\cdot \frac{1}{x^{n+1}}dx$$

Now put $1+x^{-n} = t\; $ Then $\displaystyle -\frac{1}{n}\cdot \frac{1}{x^{n+1}}dx = dt$

So $$I =-n\int\frac{1}{t}dt=-n\ln |t|+\mathcal{C} = n\ln \left|\frac{x^n}{1+x^n}\right|+\mathcal{C}$$
If you insist on doing the problem solely with substitution, you should probably try substituting $t=\sec^2\theta$. Then you end up with
$$\frac1n\int\frac{2\sec^2\theta\tan\theta\,d\theta}{\sec^2\theta\tan^2\theta} = \frac2n\int\cot\theta\,d\theta,$$
which is quite straightforward.