The definite Integral $I=\int_{0}^{1}\frac{x\cdot \operatorname{artanh} x}{1+x^2} dx$

Question

Consider:$$I=\int_{0}^{1}\dfrac{x\cdot {\operatorname{artanh} x}}{1+x^2} dx$$

Online integral calculators approximate the result to be $0.30842513753$, which is very close to $\dfrac{\pi^2}{32}$

I tried breaking the first term into an infinite series and evaluating, similar to this question:Please verify this alternate proof of the Basel problem

Now, this does result in the correct answer however, the method is long and lacks rigor.

How can I evaluate this integral in a simpler way? Is there a way to generalize similar integrals?

Answer 1

Note that $ \operatorname{artanh} x =\frac12\ln \frac{1+x}{1-x} $ and substitute $t=\frac{1-x}{1+x}$ \begin{align} &\int_{0}^{1}\dfrac{x\ {\operatorname{artanh} x}}{1+x^2}dx\\ =&\ \frac12\int_0^1 \frac{t\ln t}{1+t^2}\overset{t^2\to t}{dt}-\frac12\int_0^1 \frac{\ln t}{1+t}dt =- \frac38\int_0^1 \frac{\ln t}{1+t}dt =\frac{\pi^2}{32} \end{align} where $\int_0^1 \frac{\ln t}{1+t}dt=-\int_0^1 \frac{\ln (1+t)}{t}dt=-\frac{\pi^2}{12}$.

Answer 2

Introduce a parameter and differentiate w.r.t. it:

$$\begin{align*} I(a) &= \int_0^1 \frac{x \tanh^{-1}(ax)}{1+x^2} \, dx & [I(0)=0] \\[1ex] I'(a) &= \int_0^1 \frac{x^2}{\left(1-a^2x^2\right) \left(1+x^2\right)} \, dx \\ &= \frac1{1+a^2} \left(\int_0^1 \frac{dx}{1-a^2x^2} - \int_0^1 \frac{dx}{1+x^2}\right) \\ &= \frac1{1+a^2} \left(\frac{\tanh^{-1}a}{a} - \frac\pi4\right) \\[1ex] I(a) &= \int_0^a \left(\frac1b-\frac b{1+b^2}\right) \tanh^{-1}b \, db - \frac\pi4 \arctan a \end{align*}$$

As $a\to1^+$, we find that

$$2 I(1) = \int_0^1 \frac{\tanh^{-1}x}x \, dx - \frac{\pi^2}{16} \implies I(1) = \boxed{\frac{\pi^2}{32}}$$

where $\int_0^1\frac{\tanh^{-1}x}x\,dx=\frac{\pi^2}8$.