<s>[INST] Finding $ \int^1_0 \frac{\ln(1+x)}{x}dx$ There is supposed to be a clean solution to the integral below, maybe involving some symmetry
$$ \int^1_0 \frac{\ln(1+x)}{x}dx$$
I have tried integration by parts as followed:
$\ln(x+1)=u$ ,$\frac{1}{x+1} dx = du$ and also $\frac{1}{x}dx=dv$, $\ln(x)=v$. Then, the integral becomes
$$\int^1_0 \frac{\ln(1+x)}{x}dx=- \int^1_0 \frac{\ln x}{1+x}dx$$
which does not make this easier.
I have also tried using the identity $\int_a^bf(x)~dx=\int_a^bf(a+b-x)~dx$. So let $I = \int^1_0 \frac{\ln(1+x)}{x}dx$ and also $I = \int^1_0 \frac{\ln(2-x)}{1-x}  $. Then
$$2I= \int^1_0 \ln(1+x)+\ln(2-x)$$
which is not any easier, either.
Any ideas? :)
 [/INST] HINT:
use the expansion $$ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots$$
$$\frac{ln(1+x)}{x}=1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\ldots$$
$$\int \frac{ln(1+x)}{x}\ dx=x-\frac{x^2}{2^2}+\frac{x^3}{3^2}-\frac{x^4}{4^2}+\ldots$$
</s>