Prove that $\int_{0}^{\infty}\frac{dx}{1+x^n}=\int_{0}^{1}\frac{dx}{(1-x^n)^{1/n}}$ Prove that $\int_{0}^{\infty}\frac{dx}{1+x^n}=\int_{0}^{1}\frac{dx}{(1-x^n)^{1/n}}$,where $n>1$
My Attempt:
$\int_{0}^{1}\frac{dx}{(1-x^n)^{1/n}}=\int_{0}^{1}\frac{dx}{x(x^{-n}-1)^{1/n}}=\int_{0}^{1}\frac{x^{-n-1}dx}{x^{-n}(x^{-n}-1)^{1/n}}$
Now put $x^{-n}-1=t$
$\int_{\infty}^{0}\frac{dt}{-n(t+1)(t)^{1/n}}=\int_{0}^{\infty}\frac{dt}{n(t+1)(t)^{1/n}}=$,then i am lost how to move ahead.
 A: Substitute $x=u^{1/n}$ and $u=\frac{v}{1-v}$ and $v=t^n$:
$$
\begin{align}
\int_0^\infty\frac{\mathrm{d}x}{1+x^n}
&=\int_0^\infty\frac{\frac1nu^{\frac1n-1}\,\mathrm{d}u}{1+u}\\
&=\int_0^1\frac{\frac1n\left(\frac{v}{1-v}\right)^{\frac1n-1}\frac1{(1-v)^2}\,\mathrm{d}v}{1+\frac{v}{1-v}}\\
&=\int_0^1\left(\frac{t^n}{1-t^n}\right)^{\frac1n-1}\frac{t^{n-1}}{1-t^n}\,\mathrm{d}t\\
&=\int_0^1\frac{\mathrm{d}t}{\left(1-t^n\right)^{1/n}}\\
\end{align}
$$
A: $$\displaystyle I = \int_{0}^{1}\frac{1}{\left(1-x^n\right)^{\frac{1}{n}}}dx\;,$$ Put $\displaystyle x=\frac{1}{t}$ and $\displaystyle dx = -\frac{1}{t^2}dt$
Above we have used the formula $$\displaystyle \bullet \; \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx$$
We get $$\displaystyle I = \int_{\infty}^{1}\frac{t}{(t^n-1)^{\frac{1}{n}}}\cdot -\frac{1}{t^2}dt = \int_{1}^{\infty}\frac{1}{t\cdot (t^n-1)^{\frac{1}{n}}}dt$$
Again Above we have used the formula $$\displaystyle \bullet \; \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx$$
Now Put $t^n-1 = u^n\;,$ Then $nt^{n-1}dt = nu^{n-1}du\Rightarrow \displaystyle dt = \frac{t\cdot u^{n-1}}{u^n+1}du$
and Changing Limit
$$\displaystyle I = \int_{0}^{\infty}\frac{u^{n-1}}{u\cdot \left(1+u^n\right)}du = \int_{0}^{\infty}\frac{u^n}{u^2\left(1+u^{n}\right)}du$$
Now Put $\displaystyle u = \frac{1}{v}\;,$ Then $\displaystyle du = -\frac{1}{v^2}dv$ and Changing Limit, We get
$$\displaystyle I = -\int_{\infty}^{0}\frac{1}{v^n+1}dv = \int_{0}^{\infty}\frac{1}{1+v^n}dv = \int_{0}^{\infty}\frac{1}{1+x^n}dx$$
Again Above we have used the formula $$\displaystyle \bullet \; \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx$$ and  formula $$\displaystyle \bullet \; \int_{a}^{b}f(v)dv = \int_{a}^{b}f(x)dx$$
