The problem is: ($x≥0$ is a real number)
$$\int\sqrt\frac{\arctan(x)}{(1+x^2)}dx$$
I tried u-substitution technique where I set $u=\arctan(x)$ and therefore $dx = (1+x^2)du$. The problem is that it doesn't cancel out the $x$ since there $1+x^2$ is inside a square root.
I tried the partial integration method where I set
$
F(x) = \sqrt{\arctan(x)}
$
$
G'(x) = \sqrt\frac{1}{1+x^2}
$
And then calculated:
$
F'(x) = \frac{1}{2(1+x^2)(\sqrt{\arctan(x)})}
$
$
G(x) = \ln(\sqrt{1+x^2}+x) + C
$
And you can see where this goes, it becomes even more difficult to solve. Reverse substitution, I was unable to make that work either. How do I do this? Is there a simple trick that I'm missing here? If it gives more insight into solving this, the problem is to find volume when this is rotated about the $X$-axis at an interval $[a,b]$ where $a>0$ and $b>0$
Best Answer
When rotating about $x$-axis, if $b>a>0$, the integral to solve is $$V=\pi \int_a^b \frac{\arctan x}{x^2+1} \, dx=\frac{1}{2} \pi \left(\arctan^2 b-\arctan^2 a\right)$$