# Trouble evaluating an integral

1. Jan 15, 2009

### Nano

Indefinite Integral(arctan(x/2) / (x^2 + 4) )dx

(sorry, I can't get my word equation editor to paste it here)

I have stared at this problem for a while, but I can't figure it out. I made the triangle that has tan = x/2, and that triangle has a hypotenuse of sqrt(4 + x^2). But that doesn't seem to get me anywhere. Also, I thought using a u-substitution by u = arctan(x/2) would work, because it would get rid of the arctan, but thats not leading me anywhere either.
Could someone point me in the right direction?

2. Jan 15, 2009

### Dick

u=arctan(x/2) is correct approach. What does that make du?

3. Jan 15, 2009

### Nano

That makes du = 1/ (1 + (x^2/4)) dx
But how does that cancel out the 4 + x^2 in the denominator?
I get:
Integral(u *(1 + (x^2/4)) / 4 + x^2)du

4. Jan 15, 2009

### Dick

(4+x^2)=4*(1+x^2/4). But you are also missing a factor of (1/2) in du. Don't forget the chain rule.

5. Jan 15, 2009

### Nano

Oh!!! That's kind of tricky. Thank you for your help, I appreciate it!