Proving $I_n(a,b)=\frac{a^n}{n}-bI_{n-1}(a,b)$

Question

If $$I_n(a,b)=\int_{0}^{a} \frac{x^n}{x+b} dx $$ with $b>a>0$. How can I prove that $$I_n(a,b)=\frac{a^n}{n}-bI_{n-1}(a,b)$$

I've tried to integrate by parts with the functions $u=x^n$ and $v=\frac{1}{x+b}$ without succeeding

Answer 1

Rearrange:

$$I_n(a,b)+bI_{n-1}(a,b)=\int_0^a\frac{x^n+bx^{n-1}}{x+b}dx=\int_0^ax^{n-1}\ dx=\frac{a^n}{n}$$

Answer 2

Integrate by parts as follows

$$I_n(a,b)=\int_{0}^{a} \frac{x^n}{x+b} dx = \int_{0}^{a}\frac1n\left( \frac{x}{x+b} \right)^n d[(x+b)^n] =\frac{a^n}{n}-bI_{n-1}(a,b) $$