How to proceed this integral question? $$\int_{0}^{1} \frac{\log{(1+x)}}{x^{2}+1} \ dx $$

I tried substituting x with 1/t but couldn't find the answer. Can someone provide any hint?

As many suggested I substituted x with tan t but again I got stuck at 
$$\int_{0}^{\frac{π}{4}} \log{(1+tant)} \ dt $$

I finally solved it and got the correct answer.
Hint -

Put $\tan^{-1} x = t$

Then x = tan t.

And $\frac{1}{1 + x^2} dx = dt$
Hint -

Put $\tan^{-1} x = t$

Then x = tan t.

And $\frac{1}{1 + x^2} dx = dt$