I was trying to evaluate the integral
$$ I = \int_0^1 \frac{x \tanh^{-1}x}{3+x^2} dx$$
To solve this, I considered the family of integrals
$$ I(a) = \int_0^1 \frac{ x \tanh^{-1}ax}{3+x^2} dx$$
Note that $I(0) = 0$.
Using Leibnitz rule, we get:
$$ I'(a) = \int_0^1 \frac{(x^2+3)-(3)}{(3+x^2)(1-a^2x^2)}dx$$
$$I'(a) = \int_0^1 \frac{dx}{1-a^2x^2} + 3\cdot \int_0^1 \frac{dx}{(x^2+3)(a^2x^2-1)}$$
Multiplying and dividing by $a^2$ for the second integral, we get:
$$I'(a) = \frac{1}{2a} \cdot \ln(\frac{1+a}{1-a}) + 3\cdot a \cdot \int_0^1 \frac{adx}{(a^2x^2+3a^2)(a^2x^2-1)}$$
For the second integral, I substitute $ax = v$
$$I_0(a) = 3a\cdot \int_0^a \frac{dv}{(v^2+3a^2)(v^2-1)}$$
$$ = \frac{3a}{3a^2+1} \cdot (\int_0^a \frac{dv}{v^2-1} – \int_0^a \frac{dv}{v^2+3a^2})$$
$$I_0(a) = \frac{3a}{3a^2+1} \cdot ( \frac{1}{2}\cdot \ln(\frac{1-a}{1+a}) – \frac{1}{\sqrt3 a}\cdot \frac{\pi}{6})$$
Adding the integrals, we get:
$$I'(a) = \frac{1}{2} \cdot (\frac{1}{a} – \frac{3a}{3a^2+1}) \ln(\frac{1+a}{1-a}) – \frac{\sqrt3}{3a^2+1} \cdot \frac{\pi}{6} $$
This means that
$$ I(b) = \int_0^b I'(a) da$$
But I'm not sure how to integrate the first part of $I'(a)$ – that is, I am struggling to evaluate:
$$I_1(b) = \int_0^b \frac{1}{2} \cdot (\frac{1}{a}-\frac{3a}{3a^2+1}) \ln(\frac{1+a}{1-a}) da$$
$$I_1(b) = \int_0^b \frac{1}{2} \cdot \frac{1}{a(3a^2+1)} \ln(\frac{1+a}{1-a}) da$$
We proceed by substituting $\frac{1-a}{1+a} = t$ in the integrand and obtain:
$$I_1(b) = -\frac{1}{4}\cdot \int_{\frac{1-b}{1+b}}^1 \frac{\operatorname{ln}t\cdot (1+t)}{(t^2-t+1)\cdot (1-t)}dt$$
Can anyone help me in evaluating this specific integral? I would like to know the strategies used to evaluate integrals of such kinds.
Any help is appreciated. Thanks for reading!
Best Answer
Alternatively, reduce the integral to the familiar $\int_0^1\frac{\ln(1-x)}x dx= \frac{\pi^2}6$ as follows
\begin{align}\int_0^1\frac{x \tanh ^{-1}x}{x^2+3}\overset{x\to\frac{1-x}{1+x}}{dx} =& \ \frac14 \int_0^1\ln x\ d \left( \ln\frac{(1+x)^2}{x^2+x+1}\right)\\ \overset{ibp}= & \ \frac14 \int_0^1\frac1x \ln\frac{(1-x^2)^2}{(1-x)(1-x^3)}\ dx\\ =& \ \frac1{12} \int_0^1\frac{\ln(1-x)}x dx=\frac{\pi^2}{72} \end{align} where $\int^1_0 \frac{\ln(1-x)}{x}dx = 2\int^1_0 \frac{\ln(1-x^2)}{x}dx = 3\int^1_0 \frac{\ln(1-x^3)}{x}dx $ is used in the simplification.