We will evalute the integral \(\int_0^1 \frac{\ln(x)}{1+x} dx \) using Taylor Series expansion and switching the order of integration and summation.
Firstly, we notice that $$ \frac{1}{1+x}=\sum_{n=0}^{\infty} (-1)^n x^n $$
by taylor series expansion. This series coverges for \(x \in (0,1)\)
so we can substitute this into the integral:
$$ \int_0^1 \frac{\ln(x)}{1+x} dx = \int_0^1 \ln(x) \sum_{n=0}^{\infty} (-1)^n x^n dx $$
Further, by changing the order of integration and summation, we have $$ \int_0^1 \ln(x) \sum_{n=0}^{\infty} (-1)^n x^n dx = \sum_{n=0}^{\infty} (-1)^n \int_0^1 x^n \ln(x) dx $$
Next, we evaluate the integral \(\int_0^1 x^n \ln(x) dx\) , which, by integration by parts with \(u=\ln(x)\) and \(v=x^n+1\) , we have that
$$ \int_0^1 x^n \ln(x) dx = \frac {\ln(x)x^{n+1}}{n+1}\Big|_0^1 - \int_0^1 \frac{x^n}{n+1} dx = -\frac{x^{n+1}}{(n+1)^2}\Big|_0^1 = -\frac{1}{(n+1)^2} $$
From this, we now have that $$ \sum_{n=0}^{\infty} (-1)^n \int_0^1 x^n \ln(x) dx = \sum_{n=0}^{\infty} (-1)^n (-\frac{1}{(n+1)^2}) = \sum_{n=0}^{\infty} (\frac{(-1)^{n+1}}{(n+1)^2}) = \sum_{n=1}^{\infty} (\frac{(-1)^{n}}{n^2}) = -\frac{\pi^2}{12} $$
Thus, we have that $$ \int_0^1 \frac{\ln(x)}{1+x} dx = -\frac{\pi^2}{12} $$