Integrating $\int_0^1\frac{x\ln (1+x)}{1+x^2}dx$ with restricted techniques

Question

How does one calculate the following integral?

$$ \int_0^1\frac{x\ln (1+x)}{1+x^2}dx $$

CONTEXT: Our teacher asks us to calculate the integral using only changes of variables, integrations by parts and the following known result: $$ \int_0^1 \frac{\ln x}{x-1}\,dx=\frac{\pi^2}{6}, $$ without using complex analysis, series, differentiation under the integral sign, double integrals or special functions. Some methods not respecting the teacher's requirement are found in answers to this question.

Answer 1

For start we'll prove a result that it's going to be used later. $$\boxed{\int_0^1 \frac{\ln(1+x)}{x}dx=\frac12 \int_0^1 \frac{\ln x}{x-1}dx}=\frac12\cdot \frac{\pi^2}{6}$$ proof: $$\int_0^1 \frac{\ln x}{x+1}dx+\int_0^1 \frac{\ln x}{x-1}dx=\int_0^1 \frac{2x\ln x}{x^2-1}dx\overset{x^2\to x}=\frac12 \int_0^1 \frac{\ln x}{x-1}dx $$ $$\Rightarrow \int_0^1 \frac{\ln x}{x+1}dx=-\frac12 \int_0^1 \frac{\ln x}{x-1}dx$$ $$\int_0^1 \frac{\ln(1+x)}{x}dx=\underbrace{\ln x \ln(1+x)\bigg|_0^1}_{=0}-\int_0^1 \frac{\ln x}{1+x}dx=\frac12 \int_0^1 \frac{\ln x}{x-1}dx$$

Now back to the question. Consider the following integrals:$$I=\int_0^1\frac{x\ln (1+x)}{1+x^2}dx,\quad J=\int_0^1\frac{x\ln (1-x)}{1+x^2}dx$$

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$$I+J=\int_0^1 \frac{x\ln(1-x^2)}{1+x^2}dx\overset{x^2=t}=\frac12\int_0^1 \frac{\ln(1-t)}{1+t}dt$$ Now we will integral by parts, however we can't chose $\ln(1+t)'=\frac{1}{1+t}$ since we run into divergence issues.

We will take $(\ln (1+t)-\ln 2)'=\frac{1}{1+t}$ then: $$I+J=\frac12\bigg[\underbrace{\ln(1-t)(\ln(1+t)-\ln 2)\bigg]_0^1}_{=0}+\frac12 \int_0^1 \frac{\ln\left(\frac{1+t}{2}\right)}{1-t}dt$$ Now substitute $t=\frac{1-y}{1+y}$ in order to get: $$I+J=-\frac12 \int_0^1 \frac{\ln(1+y)}{y(1+y)}dy=\frac12 \int_0^1 \frac{\ln(1+y)}{1+y}dy-\frac12\int_0^1 \frac{\ln(1+y)}{y}dy$$ $$=\frac14 \ln^2(1+y)\bigg|_0^1 -\frac14 \cdot \frac{\pi^2}{6}=\boxed{\frac{\ln^2 2}{4}-\frac{\pi^2}{24}}$$

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Similarly, for $I-J$ set $\frac{1-x}{1+x}= t$ to get: $$I-J=-\int_0^1 \frac{x\ln\left(\frac{1-x}{1+x}\right)}{1+x^2}dx=\underbrace{\int_0^1 \frac{t\ln t}{1+t^2}dt}_{t^2\rightarrow t}-\int_0^1 \frac{\ln t}{1+t}dt$$ $$=\frac14 \int_0^1 \frac{\ln t}{1+t}dt-\int_0^1 \frac{\ln t}{1+t}dt=-\frac34 \left(\underbrace{\ln(1+t)\ln t \bigg|_0^1}_{=0} -\int_0^1 \frac{\ln(1+t)}{t}dt\right)$$ $$=\frac34 \int_0^1 \frac{\ln(1+t)}{t}dt=\frac38 \int_0^1 \frac{\ln t}{t-1}dt=\boxed{\frac{\pi^2}{16}}$$

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Finally we can extract the integral as: $$I=\frac12 \left((I+J)+(I-J)\right)=\frac12\left(\frac{\ln^2 2}{4}-\frac{\pi^2}{24}+\frac{\pi^2}{16}\right)=\boxed{\frac{\ln^2 2}{8}+\frac{\pi^2}{96}}$$ Supplementary result is following: $$\boxed{\int_0^1\frac{x\ln (1-x)}{1+x^2}dx=\frac{\ln^2 2}{8}-\frac{5\pi^2}{96}}$$

Answer 2

Alternative solution:

\begin{align} J&=\int_0^1 \frac{x\ln(1+x)}{1+x^2}\,dx\\ A&=\int_0^1 \frac{\ln(1+x^2)}{1+x}\,dx\\ &=\Big[\ln(1+x^2)\ln(1+x)\Big]_0^1-2\int_0^1 \frac{x\ln(1+x)}{1+x^2}\,dx\\ &=\ln^2 2-2J\\ J&=\frac{1}{2}\ln^2 2-\frac{1}{2}A\\ B&=\int_0^1 \frac{\ln\left(\frac{1-x^2}{1+x^2}\right)}{1+x}\,dx\\ C&=\int_0^1 \frac{\ln\left(1-x\right)}{1+x}\,dx\\ \end{align} Perform the change of variable $y=\dfrac{1-x^2}{1+x^2}$, \begin{align} B&=\frac{1}{2}\int_0^1 \frac{\left(\sqrt{1+x}-\sqrt{1-x}\right)\ln x}{x\sqrt{1-x}(1+x)}\,dx\\ &=\frac{1}{2}\int_0^1 \left(\frac{1}{\sqrt{1-x^2}}-\frac{1}{1+x}\right)\frac{\ln x}{x}\,dx\\ &=\frac{1}{2}\Big[\left(\ln(1+x)-\ln(1+\sqrt{1-x^2})+\ln 2\right)\ln x\Big]_0^1-\\ &\frac{1}{2}\int_0^1\frac{\ln(1+x)-\ln(1+\sqrt{1-x^2})+\ln 2}{x}\,dx\\ &=-\frac{1}{2}\int_0^1 \frac{\ln(1+x)}{x}\,dx+\frac{1}{2}\int_0^1\left(\frac{\ln(1+\sqrt{1-x^2})}{x}-\frac{\ln 2}{x}\right)\,dx \end{align} In the second integral, perform the change of variable $y=\dfrac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}}$, \begin{align} B&=-\frac{1}{2}\int_0^1 \frac{\ln(1+x)}{x}\,dx-\frac{1}{4}\int_0^1 \frac{(1-x)\ln(1+x)}{x(1+x)}\,dx\\ &=-\frac{1}{2}\int_0^1 \frac{\ln(1+x)}{x}\,dx-\frac{1}{4}\int_0^1\left(\frac{\ln(1+x)}{x}-\frac{2\ln(1+x)}{1+x}\right)\,dx\\ &=-\frac{3}{4}\int_0^1 \frac{\ln(1+x)}{x}\,dx+\frac{1}{4}\Big[\ln^2(1+x)\Big]_0^1\\ &=\left(-\frac{3}{4}\Big[\ln x\ln(1+x)\Big]_0^1+\frac{3}{4}\int_0^1 \frac{\ln x}{1+x}\,dx\right)+\frac{1}{4}\ln^2 2\\ &=\frac{3}{4}\int_0^1 \frac{\ln x}{1+x}\,dx+\frac{1}{4}\ln^2 2\\ \end{align} Perform the change of variable $y=\dfrac{1-x}{1+x}$, \begin{align} C&=\int_0^1 \frac{\ln\left(\frac{2x}{1+x}\right)}{1+x}\,dx\\ &=\int_0^1 \frac{\ln 2}{1+x}\,dx+\int_0^1 \frac{\ln x}{1+x}\,dx-\int_0^1 \frac{\ln(1+x)}{1+x}\,dx\\ &=\ln^2 2+\int_0^1 \frac{\ln x}{1+x}\,dx-\frac{1}{2}\ln^2 2\\ &=\int_0^1 \frac{\ln x}{1+x}\,dx+\frac{1}{2}\ln^2 2\\ B&=\int_0^1 \frac{\ln\left(\frac{(1-x)(1+x)}{1+x^2}\right)}{1+x}\,dx\\ &=C+\int_0^1 \frac{\ln(1+x)}{1+x}\,dx-A\\ &=C+\frac{1}{2}\ln^2 2-A \end{align} Therefore, \begin{align}A&=C+\frac{1}{2}\ln^2 2-B\\ &=\left(\int_0^1 \frac{\ln x}{1+x}\,dx+\frac{1}{2}\ln^2 2\right)+\frac{1}{2}\ln^2 2-\left(\frac{3}{4}\int_0^1 \frac{\ln x}{1+x}\,dx+\frac{1}{4}\ln^2 2\right)\\ &=\frac{1}{4}\int_0^1 \frac{\ln x}{1+x}\,dx+\frac{3}{4}\ln^2 2\\ &=\frac{1}{4}\left(\int_0^1 \frac{\ln x}{1-x}\,dx-\frac{2x\ln x}{1-x^2}\,dx\right)+\frac{3}{4}\ln^2 2 \end{align} In the second integral perform the change of variable $y=x^2$, \begin{align}A&=\frac{1}{4}\left(\int_0^1 \frac{\ln x}{1-x}\,dx-\frac{1}{2}\int_0^1\frac{\ln x}{1-x}\,dx\right)+\frac{3}{4}\ln^2 2\\ &=\frac{1}{8}\int_0^1\frac{\ln x}{1-x}\,dx+\frac{3}{4}\ln^2 2\\ J&=\frac{1}{2}\ln^2 2-\frac{1}{2}\left(\frac{1}{8}\int_0^1\frac{\ln x}{1-x}\,dx+\frac{3}{4}\ln^2 2\right)\\ &=\frac{1}{8}\ln^2 2-\frac{1}{16}\int_0^1\frac{\ln x}{1-x}\,dx\\ &=\boxed{\frac{1}{8}\ln^2 2+\frac{1}{96}\pi^2}\\ \end{align}