given $$ I_n=\int_{0}^{1} \frac{x^n}{x+a} dx $$
show that
$$I_0= \ln(1+a)-\ln(a) $$ $$ I_{n} = \frac{1}{n}-a I_{n-1} $$
My best try is via partial integration
$$ \int_{0}^{1} \frac{x^n}{x+a} dx = \int_{0}^{1} \frac{x}{x+a} x^{n-1} dx = \frac{1}{n} \frac{1}{a} - \int_{0}^{1} \frac{a}{(x+a)^2} \frac{1}{n} x^n dx $$
so almost the only problem is the squared term $x+a$, so I dont know how to go on, did I overlooked something or am I completly lost? thnks for any help. Greetings.
Observe that$$I_n + aI_{n-1} = \int_0^1\frac{x^n+ax^{n-1}}{x+a}dx = \int_0^1x^{n-1}dx = \frac{1}{n}$$
And for $I_0$, we have $$I_0 = \int_0^1\frac{1}{x+a}dx = \ln(1+a)-\ln(a)$$