Evaluate the integral: $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$ Compute
$$\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$$
 A: Consider:
$$I(a) = \int_0^1 \frac{\ln (1+ax)}{1+x^2} \, dx$$
than, the derivative $I'$ is equal:
$$I'(a) = \int_0^1 \frac{x}{(1+ax)(1+x^2)} \, dx = \frac{2 a \arctan x - 2\ln (1+a x) + \ln (1+x^2)}{2(1+a^2)} \Big|_0^1\\
= \frac{\pi a + 2 \ln 2}{4(1+a^2)} - \frac{\ln (1+a)}{1+a^2}$$
Hence:
$$I(1) = \int_0^1 \left(  \frac{\pi a + 2 \ln 2}{4(1+a^2)} - \frac{\ln (1+a)}{1+a^2} \right) \, da \\
2 I(1) = \int_0^1 \frac{\pi a + 2 \ln 2}{4(1+a^2)} \, da = \frac{\pi}{4} \ln 2$$
Divide both sides by $2$ and you're done.
A: Good evening,
I've got another method by putting $x=(1-t)/(1+t)$, we obtain 
$$\int_0^1\frac{\ln (x+1)}{x^2+1}dx=\int_1^0\frac{\ln\frac{2}{1+t}}{\left(\frac{1-t}{1+t}\right)^2+1}\cdot\left\{-\frac{2}{(1+t)^2}\right\}dt =\int_0^1\frac{\ln 2-\ln (1+t)}{t^2+1}\ dt.$$
You can finish easily.
A: If $(1+x)(1+y)=2$, then
$$\begin{align}
x&=\frac{1-y}{1+y}\\
1+x^2&=2\frac{1+y^2}{(1+y)^2}\\
\frac{1+x^2}{1+x}&=\frac{1+y^2}{1+y}
\end{align}\tag{1}
$$
and since $(1+y)\,\mathrm{d}x+(1+x)\,\mathrm{d}y=0$ we get
$$
\frac{\mathrm{d}x}{1+x^2}=-\frac{\mathrm{d}y}{1+y^2}\tag{2}
$$
Therefore,
$$
\begin{align}
\int_0^1\frac{\log(1+x)}{1+x^2}\mathrm{d}x
&=\int_0^1\frac{\log(2)-\log(1+y)}{1+y^2}\mathrm{d}y\tag{3}
\end{align}
$$
Adding the left side to both sides and dividing by $2$ yields
$$
\begin{align}
\int_0^1\frac{\log(1+x)}{1+x^2}\mathrm{d}x
&=\frac12\int_0^1\frac{\log(2)}{1+y^2}\mathrm{d}y\\
&=\frac\pi8\log(2)\tag{4}
\end{align}
$$
A: Start with $$\begin{align*}
\int_0^{\pi/4} \ln(1+\tan x)dx &= \int_0^{\pi/4} \ln(\sin x+\cos x)dx - \int_0^{\pi/4} \ln(\cos x)dx \\
 &= \int_0^{\pi/4} \ln\left(\cos(x-\frac{\pi}4)\right)dx +\int_0^{\pi/4} \ln(\sqrt 2)dx - \int_0^{\pi/4} \ln(\cos x)dx.
\end{align*}$$ Now change $\pi/4-x=t$ in the first integral: $$=\int_0^{\pi/4} \ln(\cos t) dt +\int_0^{\pi/4} \ln(\sqrt 2)dx - \int_0^{\pi/4} \ln(\cos x)dx$$ and the result follows. Changing $x=\tan u$ in the first integral yields your integral. As far as I know these are said Bertrand's integrals. 
@Chandrasehkar: see here http://ocw.mit.edu/courses/mathematics/18-304-undergraduate-seminar-in-discrete-mathematics-spring-2006/projects/integratnfeynman.pdf
A: Note
\begin{align}
&\int_0^1\frac{\ln (x+1)}{x^2+1}dx
=\int_0^1\underset{x\to\frac{1-x}{1+x}}{\frac{\ln \frac{x+1}{\sqrt{x^2+1}}}{x^2+1}}dx
 + \int_0^1\frac{\ln \sqrt{x^2+1}}{x^2+1}dx\\
&= \int_0^1\frac{\ln \frac{\sqrt2}{\sqrt{x^2+1}}}{x^2+1}dx
 + \int_0^1\frac{\ln \sqrt{x^2+1}}{x^2+1}dx = \frac{\ln2}2\int_0^1\frac{dx}{x^2+1}=\frac\pi8\ln2
\end{align}
A: Let us consider $$A=\iint_{[0,1]^2}\frac{x}{(1+xy)(1+x^2)}dx dy$$
By Fubini's theorem, we have : $$A=\int_0^1\left[\frac{1}{1+x^2}\int_0^1\frac{x\,dy}{1+xy}\right]dx=\int_0^1\frac{\ln(1+x)}{1+x^2}dx$$ and$$A=\int_0^1\left[\int_0^1\frac{x}{(1+xy)(1+x^2)}dx\right]dy$$But$$\frac{x}{(1+xy)(1+x^2)}=\frac{1}{1+y^2}\left(\frac{-y}{1+xy}+\frac{x+y}{1+x^2}\right)$$and therefore$$A=\int_0^1\frac{1}{1+y^2}\left(-\ln(1+y)+\frac{\ln(2)}{2}+\frac{\pi y}{y}\right)dy=-A+\frac{\pi\ln(2)}{4}$$Finally :$$\boxed{\int_0^1\frac{\ln(1+x)}{1+x^2}dx=\frac{\pi\ln(2)}{8}}$$
A: Put $x = \tan\theta$, then your integral transforms to $$I= \int_{0}^{\pi/4} \log(1+\tan\theta) \ d\theta. \tag{1}$$
Now using the property that $$\int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx,$$ we have $$I = \int_{0}^{\pi/4} \log\biggl(1+\tan\Bigl(\frac{\pi}{4}-\theta\Bigr)\biggr) \ d\theta = \int_{0}^{\pi/4} \log\biggl(\frac{2}{1+\tan\theta} \biggr) \ d\theta.\tag{2}$$
Adding $(1)$ and $(2)$ we get $$2I = \int_{0}^{\pi/4} \log(2) \ d\theta\Rightarrow I= \log(2) \cdot \frac{\pi}{8}.$$
A: You can use the residue theory to calculate
$$J=\int_{-\infty}^\infty \frac{\ln|1+x|}{(1+x^2)} \, dx $$
by putting 
$$f(z)=\frac{\ln(1+z)}{(1+z^2)}$$ and we obtain 
$$J = 2\pi i\frac{\ln(1+i)}{(i+i)}=\pi\ln\left(\sqrt2e^{\frac14\pi i}\right)=\pi\left(\frac12\ln2+\frac14\pi i\right)=\frac{\pi}{2}\ln 2+\frac{\pi^2i}4$$ 
$$J=\int_{-\infty}^{-1}\frac{\ln(1+x)}{1+x^2} \, dx 
+\int_{-1}^0 \frac{\ln(1+x)}{1+x^2} \, dx 
+\int_0^1\frac{\ln(1+x)}{1+x^2} \, dx 
+\int_1^\infty \frac{\ln(1+x)}{1+x^2} \, dx$$ 
Then by putting $(t-1)e^{\pi i}=1+x$ in the first integral we obtain:
$$\int_{-\infty}^{-1}\frac{\ln(1+x)}{1+x^2}dx=\int_1^{\infty}\frac{\pi i+\ln(t-1)}{t^2+1}dt=\frac{\pi^2i}4+\int_1^{\infty}\frac{\ln(t-1)}{t^2+1}dt$$
Then by putting $t=-x$ in the second integral we obtain:
$$\int_0^1\frac{\ln(1-x^2)}{1+x^2}dx+\int_1^{\infty}\frac{\ln(x^2-1)}{1+x^2}dx=\frac{\pi}2\ln2$$
then by putting $t=1/x$ in the second integral we obtain:
$$\int_0^1\frac{\ln(1-x^2)}{1+x^2}dx-\int_0^1\frac{\ln x}{1+x^2}dx=\frac{\pi}4\ln2$$
and we put $x=(1-t)/(1+t)$ in the first integral 
we obtain after calculus :
$$\int_0^1 \frac{\ln \, (1+x)}{1+x^2}= \frac{\pi}{8} \ln 2$$
