integrate $\int_{-1}^1 \frac{x}{2}\ln\frac{1+x}{1-x}\,dx=1$ between $-1$ and $1$.

Question

Mathematica tells me that $$\int_{-1}^1 \frac{x}{2}\ln\frac{1+x}{1-x}\,dx=1.$$

But I can't figure out how to do the integral by hand.

Would anyone care to give me a hand?

Answer 1

Set $f(x):=x ln(1+x).$

What you have to establish is that $\int_{-1}^1 (f(x)+f(-x))dx=2$

It suffices then to find an antiderivative to $f$.

It is : $F(x)=\dfrac{x^2-1}{2}\ln(x+1)-\dfrac{x^2}{4}+\dfrac{x}{2}$

Explanation

• Begin by a change of variable $u=x+1$, then $f(x)$ becomes $\varphi(u)=(u-1)\ln(u)=u \ln(u) - 1\ln(u)$

• Then integrate by parts.

In fact, one can easily find back that a primitive function of $x^n \ln(x)$ is $\dfrac{1}{n+1}\left(x^{n+1}\ln(x)-\dfrac{x^{n+1}}{n+1}\right).$

Remark: do you know that the function under the integral sign is $x ArcTanh(x) $ ?