Is it possible to integrate $\int_0^1 \frac{x^{n}}{x+5} dx + \int_0^1 \frac{x^{n-1}}{x+5} dx$? If so, how?

Question

Maybe there is some method I am forgetting, but I do not remember how to solve this (if it is even solvable). $$\int_0^1 \frac{x^{n}}{x+5} dx + \int_0^1 \frac{x^{n-1}}{x+5} dx$$

Answer 1

It is solvable, although the answer is ugly: \begin{align*}\int_0^1\frac{x^{n-1}(x+1)}{x+5}dx&=\int_0^1x^{n-1}\bigg(1-\frac 4{x+5}\bigg)dx\\ &=\frac1n-4\int_0^1 \frac{x^{n-1}}{x+5}dx\\ &=\frac1n-4\int_5^6 \frac{(u-5)^{n-1}}{u}du\\ &=\frac1n-4\int_5^6\frac1u\bigg(\sum_{r=0}^{n-1}\binom{n-1}r u^{n-r-1}(-5)^r \bigg)du\\ &=\frac1n-4\int_5^6 \sum_{r=0}^{n-2}\binom {n-1}ru^{n-r-2}(-5)^r +\frac{(-5)^{n-1}}{u}du\\ &=\frac1n-4\sum_{r=0}^{n-2}\binom {n-1}r\frac{(-5)^r (6^{n-r-1}-5^{n-r-1})}{n-r-1}-4(-5)^{n-1}\log\frac 65.\end{align*}

Answer 2

If you want a "closed-form" version without the summation, it can be written as

$$\frac{1}{n} - \frac{4}{5}\Phi\left(-\frac{1}{5},1,n\right) $$ where $\Phi$ is the Lerch Phi function.

This comes from the series

$$\frac{1}{x+5} = \frac{1}{5} \sum_{k=0}^\infty \left(-\frac{x}{5}\right)^k $$

so that $$\eqalign{\int_0^1 \frac{x^n}{x+5}\; dx &= \frac{1}{5} \sum_{k=0}^\infty \left(-\frac{1}{5}\right)^k \int_0^1 x^{n+k}\; dx = \frac{1}{5} \sum_{k=0}^\infty \frac{1}{n+k+1} \left(-\frac{1}{5}\right)^k \cr &= \frac{1}{n} - \Phi\left(-\frac{1}{5},1,n\right)}$$ and similarly $$ \eqalign{\int_0^1 \frac{x^{n-1}}{x+5} \; dx &= \frac{1}{5} \sum_{k=0}^\infty \frac{1}{n+k} \left(-\frac{1}{5}\right)^k \cr & = \frac{\Phi\left(-\frac{1}{5},1,n\right)}{5}}$$

Answer 3

Using special functions, we also could use the gaussian hypergeometric function and get for $$I_n=\int_0^1\frac{x^{n-1}(x+1)}{x+5}\,dx$$

$$I_n=\frac{n \Big(\, _2F_1\left(1,n;n+1;-\frac{1}{5}\right)+\,_2F_1\left(1,n+1;n+2;-\frac{1}{5}\right)\Big)+\, _2F_1\left(1,n;n+1;-\frac{1}{5}\right)}{5 n (n+1)}$$