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.

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: $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$
