Compute $\int_0^2\frac {\arctan{x}}{x^2+2x+2}dx$ 
Compute $I=\int_0^2\frac {\arctan{x}}{x^2+2x+2}dx$

My 2 attempts:
First:
We observe that $\frac {1}{x^2+2x+2}=\frac {1}{(x+1)^2+1}$ and $\frac 1{(x+1)^2+1}=(\arctan{(x+1)})^{'}.$ Then:
$$\int_0^2\frac{\arctan{x}}{x^2+2x+2}dx=\int_0^2\arctan{x}(\arctan{(x+1)})^{'}dx=\left.\arctan{x}\arctan{(x+1)}\right|_0^2-\int_0^2\frac{1\times\arctan{(x+1)}}{1+x^2}dx=\left.\arctan{x}\arctan{(x+1)}\right|_0^2-\int_0^2\frac{(1+x^2-x^2)\times\arctan{(x+1)}}{1+x^2}dx=\left.\arctan{x}\arctan{(x+1)}\right|_0^2-\int_0^2\arctan{(x+1)}dx-\int_0^2\frac{x^2\arctan{(x+1)}}{1+x^2}dx$$
Where the second integral is pretty easy to solve using integration by parts but the second one is not very pleasant so... I thought I should try something else.
Second attempt:
Let $u=\arctan(x+1)$ then $x=\tan u-1$ then $$I=\int_{\arctan1}^{arctan3}\arctan{(\tan u -1)du}$$
then I could let $w=\tan(u)-1$ then $dw=((w+1)^2+1)du$... but it's a lot of work and I really think you can do this more easily...
Any hints?

Some thoughts here:
\begin{align}
I&=\int\frac{\arctan x dx}{(x+1)^2+1}\\
&=\int\frac{\arctan (u-1) du}{u^2+1}\\
&=\int\frac{\arctan(\tan w-1)\sec^2w dw}{\sec^2w}\\
&=\int\arctan(\tan w-1)dw
\end{align}
Using similar substitutions, consider
$$J=\int\frac{\text{arccot} x dx}{(x+1)^2+1}=\int\text{arccot}(\tan w-1)dw$$
Notice
$$\arctan x+\text{arccot}x=\frac\pi 2$$
So
$$I+J=\frac{\pi}{2}\int^2_0\frac{dx}{(x+1)^2+1}=\frac{\pi}2(\arctan3-\frac{\pi}4)$$
