How to take $\int_0^{+\infty}\frac{\arctan x}{(1+x^2)^{\frac{3}{2}}}\;\mathrm{d}x$? The integral:
$$\int_0^{+\infty}\frac{\arctan x}{(1+x^2)^{\frac{3}{2}}}\;\mathrm{d}x$$
so there's the function and its derivative, I was thinking of substitution, but not sure which one...
 A: Letting $x=\tan t$ gives$$I=\int_{0}^{\pi/2} t \ \cos t \ \text{d}t$$You can use integration by parts now.
Another way to continue: Let $$I(a)=\int_{0}^{\pi/2} \sin (at) \ \text{d}t=\frac{1-\cos \left(\frac{\pi a}{2}\right)}{a}.$$Then
$$I'(a)=\int_{0}^{\pi/2} t \cos (at) \ \text{d}t=\frac{\text{d}}{\text{d}a} \left( \frac{1-\cos \left(\frac{\pi a}{2}\right)}{a}\right).$$Plug in $a=1$.
A: Well, when you use integration by parts:
$$\mathscr{I}:=\int_0^\infty\frac{\arctan\left(x\right)}{\left(1+x^2\right)^\frac{3}{2}}\space\text{d}x=\lim_{\text{n}\to\infty}\space\left[\frac{x\cdot\arctan\left(x\right)}{\sqrt{1+x^2}}\right]_0^\text{n}-\int_0^\infty\frac{x}{\left(1+x^2\right)^\frac{3}{2}}\space\text{d}x\tag1$$
Now, substitute:
$$\text{u}:=1+x^2\tag2$$
So, we get:
$$\mathscr{I}=\lim_{\text{n}\to\infty}\space\left[\frac{x\cdot\arctan\left(x\right)}{\sqrt{1+x^2}}\right]_0^\text{n}-\frac{1}{2}\int_1^\infty\frac{1}{\text{u}^\frac{3}{2}}\space\text{d}\text{u}=$$
$$\lim_{\text{n}\to\infty}\space\left[\frac{x\cdot\arctan\left(x\right)}{\sqrt{1+x^2}}\right]_0^\text{n}-\frac{1}{2}\cdot\lim_{\text{m}\to\infty}\space\left[-\frac{2}{\sqrt{\text{u}}}\right]_1^\text{m}=$$
$$\lim_{\text{n}\to\infty}\space\left(\frac{\text{n}\cdot\arctan\left(\text{n}\right)}{\sqrt{1+\text{n}^2}}-\frac{0\cdot\arctan\left(0\right)}{\sqrt{1+0^2}}\right)-\frac{1}{2}\cdot\lim_{\text{m}\to\infty}\space\left(-\frac{2}{\sqrt{\text{m}}}-\left(-\frac{2}{\sqrt{1}}\right)\right)=\frac{\pi}{2}-1\tag3$$
