$\int_{-1/k}^{0}\left({x^n-1\over x-1}\right)^m\left[{1-x^n\over 1-x}+{x(m+1)[1-nx^{n-1}+(n-1)x^n]\over (1-x)^2}\right]\mathrm dx$

Question

Let $(n,m)\ge1$

$$\int_{-1/2}^{0}\left({x^n-1\over x-1}\right)^m\left[{1-x^n\over 1-x}+{x(m+1)[1-nx^{n-1}+(n-1)x^n]\over (1-x)^2}\right]\mathrm dx=I\tag1$$ Using varies trial and error techniques we managed to find the closed form for $(1)$ to be $$I={[(n-1)(n-2)-(-1)^n]^{m+1}\over 2^{(n-1)(m+1)+1}}$$

We are shrugged to generalised $(1)$ for $k\ge2$

How can we find the closed form for $(2)$?

$$\int_{-1/k}^{0}\left({x^n-1\over x-1}\right)^m\left[{1-x^n\over 1-x}+{x(m+1)[1-nx^{n-1}+(n-1)x^n]\over (1-x)^2}\right]\mathrm dx=J\tag2$$ We estimated that it still have this form $$J={[F(n,k)]^{m+1}\over 2^{(n-1)(m+1)+1}}$$

Find closed form for $(1)$ was a pain staking task

Answer 1

Consider integrating by parts the integral $$\int\left(\frac{x^n-1}{x-1}\right)^{m+1}dx$$

You get $$x\left(\frac{x^n-1}{x-1}\right)^{m+1}-\int x(m+1)\left(\frac{x^n-1}{x-1}\right)^{m}\left(\frac{1-x^n+nx^n-nx^{n-1}}{(x-1)^2}\right)dx$$

Therefore the primitive of your integral is $$x\left(\frac{x^n-1}{x-1}\right)^{m+1}$$