Integrating $\frac{\log(1+x)}{1+x^2}$ 
Possible Duplicate:
Evaluate the integral: $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} dx$ 

I am a bit stuck here in evaluating the following integral:$$\int_{0}^{1}\frac{\log(1+x)}{1+x^2}\,\mathrm dx$$.Your help is appreciated.

A thought: Notice $$\int\limits_0^1 \frac{\log(1 + x)dx}{1 + x^2} = \int\limits_0^1 \frac{\log(1 + x)dx}{1 + 2x -2x + x^2} = \int\limits_0^1 \frac{\log(1 + x)dx}{(x+1)^2 - 2x } \int\limits_0^1 \frac{\log(1 + x)dx}{(x+1)^2 - 2(x+1) + 2}$$ Now put $u = x + 1$ and so youll have
$$\int\limits_1^2 \frac{\log(u)du}{u^2 - 2u + 2} = \int\limits_1^2 \frac{\log(u)du}{(u-1)^2 + 2} = \int\limits_1^2 \log(u)d[\arctan(u-1)]$$
Now use integration by parts..
