Find $\lambda$ if $\int^{\infty}_0 \frac{\log(1+x^2)}{(1+x^2)}dx = \lambda \int^1_0 \frac{\log(1+x)}{(1+x^2)}dx$ Problem : If $\displaystyle\int^\infty_0 \frac{\log(1+x^2)}{(1+x^2)}\,dx = \lambda \int^1_0 \frac{\log(1+x)}{(1+x^2)}\,dx$ then find the value of $\lambda$. 
I am not getting any clue how to proceed as if I put $(1+x^2)\,dx =t $ then its derivative is not available. Please suggest how to proceed in this. Thanks.
 A: From Evaluating $\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx$, you can obtain
$$\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx=\pi\ln 2. $$
Define
$$I(\alpha)=\int_0^{1}\frac{\ln(\alpha x+1)}{x^2+1}dx. $$
Then
\begin{eqnarray*}
I'(\alpha)&=&\int_0^{1}\frac{x}{(\alpha x+1)(x^2+1)}dx=\int_0^1\left(\frac{x+\alpha}{(\alpha^2+1)(x^2+1)}-\frac{\alpha}{(\alpha^2+1)(x^2+1)}\right)dx.\\
&=&\frac{\pi\alpha+2\ln 2-4\ln(\alpha+1)}{4(\alpha^2+1)}.
\end{eqnarray*}
Hence
\begin{eqnarray*}
I(1)&=&\int_0^1\frac{\pi\alpha+2\ln 2-4\ln(\alpha+1)}{4(\alpha^2+1)}d\alpha\\
&=&\int_0^1\frac{\pi\alpha+2\ln 2}{4(\alpha^2+1)}d\alpha-I(1)
\end{eqnarray*}
and so
$$ I(1)=\frac{1}{2} \int_0^1\frac{\pi\alpha+2\ln 2}{4(\alpha^2+1)}d\alpha=\frac{1}{8}\pi\ln 2. $$
Thus  $\lambda=8$.
A: Setting $x=\tan y,$ 
$$I=\int_0^\infty\frac{\ln(1+x^2)}{1+x^2}\ dx=\int_0^{\dfrac\pi2}\ln(\sec^2y)\ dy=-2\int_0^{\dfrac\pi2}\ln(\cos y)\ dy (\text{ as } \cos y\ge0 \text{ here})$$ 
which is available here : Evaluate $\int_0^{\pi/2}\log\cos(x)\,\mathrm{d}x$
The Right Hand Side can be found here : Evaluate the integral: $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$
