Evaluating Improper Integral of $\frac{x\arctan{x}}{(1+x^2)^2}$

[neik]

Evaluate the integral:
$$\int^{\infty}_{0}\frac{x\arctan{x}}{(1+x^2)^2}dx$$

can anybody give me some hint?
Thanks in advance

 

[Galileo]

Integrating by parts and substituting $u=\arctan x$ will work.

 

[dextercioby]

I hope u know how to choose the "u" and the "dv' for the part integration...

Daniel.

 

[neik]

I did like this:

Let $$u=\arctan{x} \Rightarrow du=\frac{dx}{1+x^2}$$

Let $$dv=\frac{x}{(1+x^2)^2}dx \Rightarrow v=-\frac{1}{2(1+x^2)}$$

$$\int{udv}=uv-\int{vdu}<br /> =-\frac{\arctan{x}}{2(1+x^2)}dx+\frac{1}{2}\int{\frac{dx}{(1+x^2)^2}}$$

and then I'm blocked here:
$$\int{\frac{dx}{(1+x^2)^2}$$

 

[dextercioby]

Use the substituion that Galileo prescribed...Denote the second integral by I:

$$I=:\int \frac{dx}{(1+x^{2})^{2}}$$

Make the substitution:

$$x=\tan u$$ and say what u get...

Daniel.