Compute $\int\frac{1}{(1+x^n)\sqrt[n]{1+x^n}}dx$

Question

How to compute this integral:$$\int\frac{1}{(1+x^n)\sqrt[n]{1+x^n}}dx$$I tried to make$$\ t^n = 1+x^n$$ But I got a more complicated formula$$\int\frac{1}{t^{n+1}}\frac{t^{n-1}}{{(t^n-1)}^\frac{n-1}{n}}dt$$then I can not go on

Answer 1

$$t=(1+x^n)^{-1/n}\\dt=-\frac{x^{n-1}}{(1+x^n)^{-(n+1)/n}}dx\\t^{-n}=1+x^n\\x^{n-1}=(t^{-n}-1)^{(n-1)/n}\\x^{n-1}=\frac{(1-t^n)^{(n-1)/n}}{t^{n-1}}\\\int-\frac{t^{n-1}}{(1-t^n)^{(n-1)/n}}dt\\(1-t^n)^{1/n}=c\\-\frac{t^{n-1}}{(1-t^n)^{(n-1)/n}}dt=dc\\\int dc=c+C=(1-t^n)^{1/n}=(1-(1+x^n)^{-1})^{1/n}=(1-\frac{1}{1+x^n})^{1/n}=(\frac{x^n}{1+x^n})^{1/n}+C=x(1+x^n)^{-1/n}+C$$

Answer 2

It is equal to $x{(1+x^n)}^{-1/n}+C$.

Answer 3

HINT:

Choose $x=1/y$

$\implies dx=-\dfrac{dy}{y^2}$

and $\dfrac1{(1+x^n)^{1+1/n}}=\dfrac{y^{n+1}}{(y^n+1)^{1+1/n}}$

Now set $y^n+1=u$

Or directly, $u=y^n+1=\dfrac{x^n}{1+x^n}$