Find $\int_0^1\ln\left(\frac{1+x}{1-x}\right)~dx$ without resorting to series 
Find $$\int_0^1\ln\left(\frac{1+x}{1-x}\right)~dx$$

Firstly, note that the upper limit makes this an improper integral.
The method I employed against this problem was integration by parts, as shown below:
$$\int_0^1\ln\left(\frac{1+x}{1-x}\right)~dx=2\int_0^1\text{artanh} x ~dx=2\left[x~\text{artanh}~x+\frac{1}{2}\ln\lvert 1-x^2\rvert\right]_0^1$$
However, I am unsure how to find the value that this converges to.
I then tried to find a series representation of the value of the integral using the Maclaurin expansion of $\text{artanh}~x$, which was successful; I found that the value of the integral is
$$2\sum_{r=0}^\infty\frac{(2r)!}{(2r+2)!}=2\sum_{r=0}^\infty\left(\frac{1}{2r+1}-\frac{1}{2r+2}\right)=2\sum_{r=0}^\infty\frac{(-1)^r}{r+1}$$
which I know is $2\ln2$, which is the answer.
However, I would like to know how to use a regular method of integration such as my first method and still calculate the limit it converges to.
Thank you for your help.
 A: Just to give a different approach, first substitute $u={1+x\over1-x}$ and then integrate by parts:
$$u={1+x\over1-x}\implies x={u-1\over u+1}\implies dx={2du\over(u+1)^2}$$
so
$$\begin{align}
\int_0^1\ln\left(1+x\over1-x\right)\,dx
&=2\int_1^\infty{\ln u\over(u+1)^2}\,du\\
&={-2\ln u\over(u+1)}\Big|_1^\infty+2\int_1^\infty{1\over u(u+1)}\,du\\
&=(0-0)+2\int_1^\infty\left({1\over u}-{1\over u+1}\right)\,du\\
&=2\ln\left(u\over u+1\right)\Big|_1^\infty\\
&=2(\ln1-\ln(1/2))\\
&=2\ln2
\end{align}$$
A: Simplify the limits by integrating by parts as follows
\begin{align}
\int_0^1\ln\left(\frac{1+x}{1-x}\right)~dx
&= -2\int_0^1 \tanh^{-1}x\> d(1-x)\\
&= -2(1-x) \tanh^{-1}x\bigg|_0^1+2 \int_0^1 \frac{dx}{1+x}
=0+2\ln2
\end{align}
A: You already found the indefinite integral as $$2x\operatorname{arctanh}\left(x\right)+\ln\left(1-x^{2}\right)$$ Evaluated at $x = 0$, it is clearly $0$. This function can be rewritten as $$\left(1+x\right)\ln\left(1+x\right)+\left(1-x\right)\ln\left(1-x\right),$$ so the answer is $$\lim_{x \to 1}\left( \left(1+x\right)\ln\left(1+x\right)+\left(1-x\right)\ln\left(1-x\right) \right) = 2\ln(2)+\lim_{x \to 1}\left(1-x\right)\ln\left(1-x\right)$$
The last limit is equivalent to $\lim_{x \to 1}\frac{\ln\left(1-x\right)}{\frac{1}{1-x}}$, which by L'Hopital is $\lim_{x \to 1}(x-1) = 0$. Therefore, the final integral is $$2\ln(2) + 0 = 2\ln(2)$$
A: Begin by doing the sub $x \mapsto 1-x$. You have
$$\int_0^1 \log\left({1+x\over 1-x}\right) = \int_0^1 \log\left({1+(1-x)\over x}\right) = \int_0^1 \log(2-x)\,dx  - \int_0^1 \log(x)\,dx$$
Finish by using integration-by-parts.
