Find $\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}$

Question

Find $$\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}$$

with $n \in \mathbb{N}$.

My tried: I think that, I need to find the value of

$$I_1=\int_{0}^1 \frac{dx}{(1+x^n)\sqrt[n]{1+x^n}}$$

because:

$$\begin{align} I_1=\int_{0}^1 \frac{dx}{(1+x^n)\sqrt[n]{1+x^n}} &= \frac{x}{(1+x^n)\sqrt[n]{1+x^n}} \Bigg|_0^1-\int_0^1 x \ d(1+x^n)^{-1-\frac{1}{n}} \\ &=\frac{1}{2\sqrt[n]{2}}-\int_0^1 x\left(-1-\frac{1}{n}\right)(1+x^n)^{-2-\frac{1}{n}}(nx^{n-1}) \ dx \\ &=\frac{1}{2\sqrt[n]{2}}-\int_0^1(-n-1)\frac{x^n}{(1+x^n)^2\sqrt[n]{1+x^n}} \ dx\\ &=\frac{1}{2\sqrt[n]{2}}+(n+1)\int_0^1\frac{x^n+1-1}{(1+x^n)^2\sqrt[n]{1+x^n}} \ dx \\ &=\frac{1}{2\sqrt[n]{2}}+(n+1)I_1-(n+1)\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}.\end{align} $$

So that,

$$ (n+1)\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}=\frac{1}{2\sqrt[n]{2}}+nI_1$$

But how to find the following integral ?

$$I_1=\int_{0}^1 \frac{dx}{(1+x^n)\sqrt[n]{1+x^n}}$$

Answer 1

Observe that $$ \left(\frac{1}{(1+x^n)^{1/n}}\right)'=-\frac{x^{n-1}}{(1+x^n)^{1+1/n}} \tag1 $$ giving $$ \begin{align} \left(\frac{x}{(1+x^n)^{1/n}}\right)'=\left(x \times\frac{1}{(1+x^n)^{1/n}}\right)'=\frac{1}{(1+x^n)^{1/n}}-\frac{x \times x^{n-1}}{(1+x^n)^{1+1/n}}=\frac{1}{(1+x^n)^{1+1/n}} \end{align} $$ then $$ \begin{align} I_1=\int_{0}^1 \frac{dx}{(1+x^n)\sqrt[n]{1+x^n}}=\int_{0}^1 \frac{dx}{(1+x^n)^{1+1/n}}=\left[\frac{x}{(1+x^n)^{1/n}}\right]_{0}^{1}=\frac{1}{2^{1/n}} \end{align} $$ and you deduce the value of your initial integral:

$$ \int_{0}^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}=\frac{2n+1}{2(n+1)}\frac{1}{2^{1/n}} .\tag2 $$

Answer 2

Given $$I = \int_{0}^{1}\frac{1}{(1+x^n)^2\sqrt[n]{1+x^n}}dx\;,$$ Now Put $\displaystyle x = (\tan \theta )^{\frac{2}{n}}\;,$ Then $\displaystyle dx = \frac{2}{n}(\tan \theta)^{\frac{2}{n}-1}\cdot \sec^2 \theta d\theta$ and changing limit, We get

$$I = \frac{2}{n}\int_{0}^{\frac{\pi}{4}}\frac{(\tan \theta)^{\frac{2}{n}-1}\cdot \sec^2 \theta}{\sec^4 \theta \cdot \sec ^{\frac{2}{n}}\theta}d\theta = \frac{2}{n}\int_{0}^{\frac{\pi}{4}}\frac{(\tan \theta)^{\frac{2}{n}-1}}{(\sec \theta)^{\frac{2}{n}+2}}d\theta$$

So we get $$I = \frac{2}{n}\int_{0}^{\frac{\pi}{4}}\frac{(\sin \theta)^{\frac{2}{n}}}{\tan \theta \cdot \sec^2 \theta}d\theta = \frac{2}{n}\int_{0}^{\frac{\pi}{4}}\cos^3 \theta \cdot (\sin \theta)^{\frac{2}{n}-1}d\theta$$

So we get $$I = \frac{2}{n}\int_{0}^{\frac{\pi}{4}}\left[(\sin \theta)^{\frac{2}{n}-1}-(\sin \theta)^{\frac{2}{n}+1}\theta)\right]\cdot \cos \theta d\theta $$

So $$I = \frac{2}{n}\left[\frac{(\sin \theta)^{\frac{2}{n}}}{\frac{2}{n}}-\frac{(\sin \theta)^{\frac{2}{n}+2}}{\frac{2}{n}(n+1)}\right]_{0}^{\frac{\pi}{4}} = \left(\frac{1}{2}\right)^{\frac{1}{n}+1}\cdot \frac{2n+1}{n+1}$$

Answer 3

Calculation of $$I = \int\frac{1}{(1+x^n)\sqrt[n]{1+x^n}}dx\;,$$ Now Put $(1+x^n) = t^n\;,$ Then $nx^{n-1}=nt^{n-1}dt\Rightarrow x^{n-1}dx = t^{n-1}dt$

So we get $$I=\int\frac{(t^n-x^n)}{t^{n+1}}dx = \int \left[\frac{dx}{t}-\frac{dt}{t^2}\right] = \int d\left(\frac{x}{t}\right)=\frac{x}{t}+\mathcal{C}$$

So we get $$\int_{0}^{1}\frac{1}{(1+x^n)\sqrt[n]{1+x^n}}dx = \left[\frac{x}{\sqrt[n]{1+x^n}}\right]_{0}^{1} = \frac{1}{2^{\frac{1}{n}}}$$