Integration $I_n=\int_{0}^{1}\frac{dx}{(x^n+1)(\sqrt[n]{x^n+1})}$

Question

$$I_n=\int_{0}^{1}\frac{dx}{(x^n+1)\large\sqrt[n]{\normalsize x^n+1}}$$

Could someone help me through this problem?

Answer 1

Let $u = x^n + 1$, so that $dx = du/(nx^{n-1})$.

$$\int_0^1{\frac{dx}{(x^n+1)\sqrt[n]{x^n+1}}} = \int_1^2{\frac{du}{nx^{n-1}(u\sqrt[n]{u})}} = \frac{1}{n}\int_1^2{\frac{du}{u\sqrt[n]{u}\sqrt[n]{(u - 1)^{n-1}}}}$$ $$= \frac{1}{n}\int_1^2{\frac{du}{u^{(n+1)/n}(u - 1)^{(n - 1)/n}}}$$

I believe this last integral can be resolved using partial fractions. I'll stop here.

Answer 2

Let $\displaystyle v=v(x)=\left(\frac{x^n+1}{x^n}\right)^{\Large \frac{1}{n}}$.

Then we have: $\displaystyle v(0)=+\infty \ , \ v(1)=2^{\large \frac{1}{n}} \ , \ \frac{1}{x^n+1} = 1 - v^{-n} \ , \ x=\frac{1}{\left(v^n-1\right)^{\large \frac{1}{n}}}$ and $$\displaystyle dx = - \frac{v^{n-1}}{\left(v^n-1\right)^{1+\large \frac{1}{n}}} dv$$

Thus,

$$\displaystyle \int_0^1{\frac{dx}{(x^n+1)\large \sqrt[n]{x^n+1}}} = \int_{2^{\large \frac{1}{n}}}^{+\infty} \left(1 - v^{-n}\right)^{\large\frac{n+1}{n}} \frac{v^{n-1}}{\left(v^n-1\right)^{1+\large \frac{1}{n}}} dv =\int_{2^{\large \frac{1}{n}}}^{+\infty} \frac{dv}{v^2} = 2^{\large -\frac{1}{n}}$$

Please doublecheck my working!

Answer 3

$$\displaystyle K=\int{\frac{dx}{(x^n+1)\large\sqrt[n]{x^n+1}}}$$ 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, $$\displaystyle I=\frac{1}{\large \sqrt[n]{2}}$$