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

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

Could someone help me through this problem?

• Hint: Set $u = x^n + 1$. Then $du = nx^{n-1}dx$. Oct 23, 2014 at 11:49
• Why "complex analysis" tag?
– user170039
Oct 23, 2014 at 12:02
• @JoseArnaldoDris: Can you help me calculate it ? Sorry for my bad English :( Oct 23, 2014 at 12:12
• @GINKO: if you like me solution, could you "tick" and upvote it, pretty please? :) Oct 23, 2014 at 14:04

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,

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

• Nice solution, I edited your post a little. (+1) Oct 23, 2014 at 14:36

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

• Nice solution, I edited your post a little. (+1) Oct 23, 2014 at 14:32

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.

• No it's not possible, and it doesn't change the problem apart from reformuling it since what you have is not polynomials, so partial fractions are not accurate nor usable Oct 23, 2014 at 12:45
• @Jose: +1, helped me to solve it, even though I needed one more substitution. Oct 23, 2014 at 13:51