# Homework Help: Improper Integral

1. Mar 7, 2005

### 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

2. Mar 7, 2005

### Galileo

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

3. Mar 7, 2005

### dextercioby

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

Daniel.

4. Mar 7, 2005

### neik

I did like this:

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

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

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

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

5. Mar 7, 2005

### 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.

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