Use of substitution into the integral $\int_0^2 \frac{\arctan x}{x^2+2x+2}dx$ This question:Integral $\int_0^2 \frac{\arctan x}{x^2+2x+2}dx$ got some attention, but I didn't understand anything of it. Now I looked at the solution of the book and there is given the hint to use the change of variable $$x=\frac{1-t}{1+2t}$$ Could you please explain how one could think of such substitution? Is there some intuition behind this? Also after I used the substitution I got: $$\int_{-1/5}^1\frac{3\arctan\left(\frac{1-t}{1+2t}\right)}{5t^2+8t+5}dt$$ How can I finish this using the hint from the book?
 A: I suspect that there is a typo in the hint of your book. Indeed, consider the substitution 
$$ x \mapsto \frac{2-x}{1+2x}. \tag{*} $$
If $I$ denotes the integral, then the above substitution yields
$$ I = \int_{0}^{2} \frac{\arctan\left( \frac{2-x}{1+2x} \right)}{x^2 + 2x + 2} \, dx $$
Then by using the identity $\arctan(x)+\arctan\left( \frac{2-x}{1+2x} \right) = \arctan(2)$, we obtain
\begin{align*}
2I
= I + I
&= \int_{0}^{2} \frac{\arctan(x)+\arctan\left( \frac{2-x}{1+2x} \right)}{x^2 + 2x + 2} \, dx \\
&= \int_{0}^{2} \frac{\arctan(2)}{x^2 + 2x + 2} \, dx \\
&= \arctan(2)\arctan(1/2).
\end{align*}
Therefore $I = \frac{1}{2}\arctan(2)\arctan(1/2)$.
Remark. This is essentially the same solution as in my previous answer. After I wrote up the previous answer, I realized that I can directly work with the form given in the problem, so I wrote up this answer.
A: $$x^2+2x+2=(\frac{1-t}{1+2t})^2+2(\frac{1-t}{1+2t})+2=\frac{(1-t)^2+(1+2t)(2-2t)+2(1+2t)^2}{(1+2t)^2}=\frac{5+8t+5t^2}{(1+2t)^2}$$
$$\therefore I=\int_{-\frac{1}{5}}^{1}{\frac{arctan(\frac{1-t}{1+2t})}{5t^2+8t+5}}dt$$
Don't see how this helps as the quadratic has no real roots
