<s>[INST] Calculating $\int_0^1 \frac{x\ln(x+1)}{x^2+1} dx$ without using complex numbers One can find the antiderivate with help of the partial fraction method introducing complex numbers:
$$\frac{x\ln(x+1)}{x^2+1}=\frac{1}{2}\left(\frac{\ln(x+1)}{x-i}+\frac{\ln(x+1)}{x+i}\right).$$
The result is a complicated function containing dilogarithms and logarithms with complex arguments. The value of the definite integral turns out to be
$$\int_0^1 \frac{x\ln(x+1)}{x^2+1} dx = \frac{1}{96}(\pi^2+12\ln^2(2)).$$
Is there a way to find that value more directly - without introducing complex numbers?
 [/INST] Rewrite the integral as
$$I=\int_0^1 \frac{x\log (x+1)}{x^2+1}dx =\int_0^1\int_0^1\frac{x^2\:dydx}{(yx+1)(x^2+1)}$$
A partial fraction and integration order reverse later we get
$$\int_0^1\int_0^1 \frac{xy}{(y^2+1)(x^2+1)}-\frac{1}{(y^2+1)(x^2+1)}   +\frac{1}{xy+1}-\frac{y^2}{(y^2+1)(xy+1)}\:dxdy = \int_0^1\frac{y}{2(y^2+1)}\log(2) -\frac{\pi}{4}\frac{1}{y^2+1}+ \frac{\log(1+y)}{y}-\frac{y\log(y+1)}{y^2+1}\:dy$$
The first two terms simplify easily and the last term is the integral we started with. The third term is given by Taylor series
$$\sum_{n=1}^\infty \frac{(-1)^n}{n}\int_0^1x^{n-1}dx = \sum_{n=1}^\infty \frac{(-1)^n}{n^2} =\frac{\pi^2}{12}$$
thus the equation simplifies to
$$\implies I = \frac{\log^22}{4}-\frac{\pi^2}{16}+\frac{\pi^2}{12}- I$$
therefore $I = \frac{1}{96}(\pi^2 +12\log^22)$
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