Calculate the following integral:
$$ \int^3_{-3}\left(\frac{\arctan(\sqrt{|x|})}{1 + (1+x^2)^x} \right)dx $$ I know that each function can be represented as a sum of even and odd function such that: $$ f(x)=\frac{f(x) + f(-x)}{2} +\frac{f(x) - f(-x)}{2} $$ I tried using this approach and separate the function but I'm mainly confused with the evenness and oddness of the numerator of the function. If my approaches is correct, how can I properly separate the function?
Let $u=-x\implies du=-dx$: $$I=-\int_3^{-3} \frac{\arctan\sqrt{|u|}}{1+(1+u^2)^{-u}} dx \\ = \int_{-3}^3 \frac{(1+u^2)^u \arctan\sqrt{|u|}}{1+(1+u^2)^u} du$$ And now comes the ‘Add $I$ to itself’ trick: $$I+I = \int_{-3}^3 \frac {\arctan\sqrt{|x|}}{1+(1+x^2)^x} dx + \int_{-3}^3 \frac{(1+x^2)^x\arctan\sqrt{|x|}}{1+(1+x^2)^x} dx \\ = \int_{-3}^3 \arctan\sqrt{|x|} dx \\ =2\int_0^3 \arctan \sqrt x dx\\ = 2x\arctan \sqrt x \big |_0^3 -\int_0^3 \frac{\sqrt x}{x+1} dx \\ \overset{x=u^2}=6\arctan\sqrt 3 -2\int_0^{\sqrt 3} \frac{u^2+1-1}{u^2 +1} du \\ =6\arctan\sqrt 3 -2 \big[ u-\arctan u \big]_0^{\sqrt 3}\\ = \frac{8\pi}{3}-2\sqrt 3 \\ \implies I =\frac{4\pi}{3} -\sqrt 3 $$
In general, this technique can be used for doing complicated integrals like $\int_a^b f(x) dx $ such that $f(x) +f(a+b-x)$ can be easily integrated.
The following is always true
$$\int_{-a}^a \frac{\operatorname{even}(x)}{1+\operatorname{even}(x)^{\operatorname{odd}(x)}}dx = \int_0^a\operatorname{even}(x)dx$$
