what is the indefinite integral of $\int \frac{x^n}{x+1}\ dx$ where $n \in \mathbb{N}$

Question

I'm trying to solve $\int \frac{x^n}{x+1}\ dx$ for $n \in \mathbb{N}$.

I tried a couple of things like trigonometric substitution and by parts but somehow it didn't help much.

Answer 1

Just substitute $y=x+1$ and use the binomial formula:

$$\int \frac{x^n}{1+x}dx = \int \frac{(y-1)^n}{y}dy $$ $$= \int \frac 1y \sum_{k=0}^n(-1)^{n-k}\binom nk y^{k}dy$$ $$ = \int\frac{(-1)^n}{y}dy + \sum_{k=1}^n\int(-1)^{n-k}\binom nk y^{k-1}dy$$ $$= (-1)^n\ln |y| + \sum_{k=1}^n(-1)^{n-k}\frac{1}{k}\binom nk y^{k} (+C)$$ $$= (-1)^n\ln |x+1| + \sum_{k=1}^n(-1)^{n-k}\frac{1}{k}\binom nk (x+1)^{k} (+C)$$

Answer 2

$$ I_n = \int \frac{x^n}{x+1}dx = \int \frac{x^{n-1}(x+1)}{x+1} dx - I_{n-1} =\frac{x^n}{n} - I_{n-1} $$

etc., with $I_0 = \log (x + 1)$

Answer 3

$\int \frac {x^n}{x+1} dx$

$x = u-1\\ dx = du$

$\int \frac {(u-1)^n}{u} du$

$\int \frac {(-1)^n}{u} + \sum_\limits{k=1}^n {n\choose n-k+1}(-1)^{n-k+1} u^{n-k}\ du\\ (-1)^n \ln (x+1) + \sum_\limits{k=0}^{n-1} {n\choose n-k}(-1)^{n-k} \frac {(x+1)^{n-k}}{n-k}$