TITLE: Integration $I_n=\int_{0}^{1}\frac{dx}{(x^n+1)(\sqrt[n]{x^n+1})}$
QUESTION [6 upvotes]: $$I_n=\int_{0}^{1}\frac{dx}{(x^n+1)\large\sqrt[n]{\normalsize x^n+1}}$$
Could someone help me through this problem?

REPLY [3 votes]: $$\displaystyle K=\int{\frac{dx}{(x^n+1)\large\sqrt[n]{x^n+1}}}$$ Let $$x^{n}=t \Rightarrow x=\large\sqrt[n]{t}\Rightarrow \normalsize dx=\frac{dt}{n\large\sqrt[n]{t^{n-1}}}$$ Hence,  $$\displaystyle K=\int\frac{dt}{nt\large\sqrt[n]{\frac{t+1}{t}}}$$
Let $$\displaystyle \large \sqrt[n]{\frac{t+1}{t}}=\normalsize u\Rightarrow t=\frac{1}{u^{n}-1}\Rightarrow dt=\frac{-nu^{n-1}du}{(u^{n}-1)^{2}}$$
$$\displaystyle K=-\int\frac{du}{u^{2}}\Rightarrow K=\frac{1}{u}=\frac{x}{\large \sqrt[n]{x^{n}+1}}$$
Therefore, $$\displaystyle I=\frac{1}{\large \sqrt[n]{2}}$$