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?
 A: An elementary solution. Let $I$ denote the integral. Apply the substitution $x=\frac{2t}{t+\sqrt{5}}$ to obtain
$$ I = \int_{0}^{\infty} \frac{\arctan\left(\frac{2t}{t+\sqrt{5}}\right)}{\sqrt{5}+4t+\sqrt{5}t^2} \, dt. \tag{1} $$
Substituting $t \mapsto 1/t$, we find that
$$ I = \int_{0}^{\infty} \frac{\arctan\left(\frac{2}{1+\sqrt{5} t}\right)}{\sqrt{5}+4t+\sqrt{5}t^2} \, dt. \tag{2} $$
But it is easy to check that
$$ \arctan\left(\frac{2t}{t+\sqrt{5}}\right) + \arctan\left(\frac{2}{1+\sqrt{5} t}\right) = \arctan(2) $$
holds, either by utilizing the addition formula for arctan or by differentiating the LHS to check that the LHS is constant and then plugging $t=0$ to determine the value of the constant.
Therefore by averaging $(1)$ and $(2)$, we obtain
$$ I
= \frac{\arctan(2)}{2} \int_{0}^{\infty} \frac{dt}{\sqrt{5}+4t+\sqrt{5}t^2}
= \frac{\arctan(2)\arctan(1/2)}{2}. $$
A: I agree with Sangchul Lee. By writing $\arctan(x)$ as $\text{Im}\,\log(1+ix)$ and by using partial fraction decomposition and integration by parts one gets
$$ \int_{0}^{2}\frac{\arctan(x)}{x^2+2x+2}\,dx = -\frac{\pi^2}{48}+\frac{1}{2}\arctan(2)\arctan\left(\frac{1}{2}\right)-\frac{\log^2(5)}{8}+\frac{1}{2}\text{Re}\left[\text{Li}_2(i-2)+\text{Li}_2\left(\frac{i+2}{5}\right)\right].$$
On the other hand, by the functional relations for $\text{Li}_2$ the above line simplifies into
$$ \frac{1}{2}\arctan(2)\arctan\left(\frac{1}{2}\right) $$
which is the only reasonable option, since the given integral is clearly pretty close to one fourth.
A: 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?
