Variation of Putnam Integral I need help to evaluate the integral of $\dfrac{\ln x}{x^2 + 1}$ from $0$ to $1.$
The Putnam exam once had the same problem but with $\ln (x+1).$ Feynman's technique didn't help me once that $+1$ was gone.
What should I do?
 A: $$\begin{align}\int_{0}^{1}{\frac{\ln{\left(x\right)}}{1+x^2}dx}=\sum_{n=0}^{\infty}\left(-1\right)^n\int_{0}^{1}{\ln{\left(x\right)}x^{2n}dx}&\buildrel{{IBP}}\over{\overbrace{=}}\sum_{n=0}^{\infty}\left(-1\right)^n\left[{\underbrace{\frac{x^{2n+1}\ln{\left(x\right)}}{2n+1}}_{0}}-\frac{x^{2n+1}}{\left(2n+1\right)^2}\right]_0^1\\&=-\sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{\left(2n+1\right)^2}=-G\end{align}$$
G stands for Catalan’s Constant: https://en.wikipedia.org/wiki/Catalan%27s_constant
A: Hint: Integration by parts with $f=\ln(x)$ and $g'=1/(x^2+1)$ $$\int _0^1\frac{\ln \left(x\right)}{x^2+1}dx=\left[\ln \left(x\right)\arctan \left(x\right)-\int \frac{\arctan \left(x\right)}{x}dx\right]^1_0$$
You must consider that the 2nd integral is not elementary.
$$\int \frac{\arctan \left(x\right)}{x}dx=\frac{1}{2}i\left(\text{Li}_2\left(-ix\right)-\text{Li}_2\left(ix\right)\right)$$
A: With regard to the antiderivative, use partial fraction decomposition and write
$$\frac 1 {x^2+1}=\frac 1 {(x+i)(x-i)}=\frac i 2 \left(\frac 1 {x+i}-\frac 1 {x-i }\right)$$ and you face two integrals
$$I=\int \frac {\log(x)}{x+a}\,dx=\text{Li}_2\left(-\frac{x}{a}\right)+\log (x) \log \left(1+\frac{x}{a}\right)$$ which make
$$J(x)=\int\frac {\log(x)} {x^2+1}=\log (x) \tan ^{-1}(x)+\frac{i}{2} \left(\text{Li}_2(i x)-\text{Li}_2(-i x)\right)$$ and
$$J(1)-J(0)=-C$$
A: $$\int_0^1\frac{\ln^nx}{1+x^2}dx=\sum_{k=0}^\infty(-1)^k\int_0^1x^{2k}\ln^nx\ dx=\sum_{k=0}^\infty(-1)^k\left[\frac{(-1)^{n}n!}{(2k+1)^{n+1}}\right]$$
$$=(-1)^nn!\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^{n+1}}=(-1)^nn!\beta(n+1)$$
where $\beta(n+1)$ is the Dirichlet beta function.

To calculate $\int_0^1\frac{\ln(1+x)}{1+x^2}dx$ just set $\frac{1-x}{1+x}=y$, you will have a symmetry and the answer will be extracted right away.
