Evaluating the integral $\int_0^1\frac1{(1+x^3)^{n}}dx$.

Question

I proved the part of the equation which asked for $$ \frac{d}{dx}\left[x\cdot (1+x^3)^{-n} \right]=-(3n-1)\cdot (1+x^3)^{-n}+3n\cdot (1+x^3)^{-n-1} $$ d/dx[x(1+x)−nx(1+x)−n]= another equation. However, the last part, relating to I_n, is quite difficult to prove. I tried doing it with Integration by parts but never reached closer to the answer. Hope you provide with the full solution for the last part of the question highlighted as (i). Thank you! The link to the picture's below. P.S. I am learning MathJax formatting that's why I uploaded a picture. Hope you understand.(https://i.stack.imgur.com/QfKt5.jpg)

Answer 1

We set $$ I_n:=\int_0^1\frac1{(1+x^3)^{n}}dx. $$

Hint. You have already proved that $$ \frac{d}{dx}\left[x\cdot (1+x^3)^{-n} \right]=-(3n-1)\cdot (1+x^3)^{-n}+3n\cdot (1+x^3)^{-n-1} $$ then integrating from $0$ to $1$ gives $$ \int_0^1\left(\frac{d}{dx}\left[x\cdot (1+x^3)^{-n} \right]\right)dx=-(3n-1)\cdot \int_0^1(1+x^3)^{-n}dx+3n\cdot \int_0^1(1+x^3)^{-n-1}dx $$ or$$ \underbrace{\int_0^1\left(\frac{d}{dx}\left[x\cdot (1+x^3)^{-n} \right]\right)dx}_{\Large\color{red}{?}}=-(3n-1)\cdot I_n+3n\cdot I_{n+1} $$ Can you finish it?