Integral $\int_0^2 \frac{\arctan x}{x^2+2x+2}dx$ I am tring to evaluate
$$I=\int_0^2 \frac{\arctan x}{x^2+2x+2}dx$$
The first thing I did was to notice that
$$\frac{1}{x^2+2x+2}=\frac{1}{(x+1)^2+1}=\frac{d}{dx}\arctan(x+1)$$
So I integrated by parts in order to get
$$I=\arctan 2\arctan 3-\int_0^2\frac{\arctan(x+1)}{1+x^2}dx$$
I let $x=u+1$ but when I do that I get
$$I=\arctan 2\arctan 3+\int_{-1}^1\frac{\arctan(u)}{1+(1+u)^2}du
   =\arctan 2\arctan 3$$
Now this is not close to the approximation given by wolfram. What have I done wrong and how to solve this?

There is no need to evaluate the integral to answer the original question. 
 The original question is a multiple choice question so ruling out every option but the right one leads to the right answer of course. One sees that the integrand is positive almost everywhere. Moreover  the arctangent function is increasing, so one has:
\begin{align}
0 < \int^2_ 0 \frac{\arctan(x)}{x^2+2x+2}\,dx &\leq \arctan(2) \int^2 _0 \frac{1}{x^2+2x+2}\,dx\\&= \arctan(2)\left( \arctan(3)-\arctan(1)\right)
\end{align}
By the addition formula for arctangent function one sees that:
\begin{align}
\arctan(3)-\arctan(1) = \arctan\left( \frac 1 2 \right)
\end{align}
Now define for $x>0$ the function:
$$f(x): = \arctan(x)\arctan\left( \frac 1 x\right)$$ 
This function is strictly positive. Moreover it tends to zero as $x\to 0^+$ and as $x\to\infty$ and it is  differentiable with only one stationary point $x=1$ (check this!) which clearly corresponds with the maximum. So:
\begin{align*}
f(x) \leq f(1) = \arctan(1)^2 = \frac{\pi^2}{4^2} <  \frac{4^2}{4^2}=1
\end{align*}
Hence:
\begin{align}
0 < \int^2_ 0 \frac{\arctan(x)}{x^2+2x+2}\,dx < 1 < \pi < 2\pi
\end{align}
So what is the only option that can be the right answer?
