# Homework Help: Undefined Integral

1. Nov 24, 2012

### ruiwp13

1. The problem statement, all variables and given/known data

Solve the following integral:

∫arctg(1/√x)dx

2. Relevant equations

∫u'/1+u^2 = arctg(u)+c
∫u.v' = u.v - ∫ u'.v

3. The attempt at a solution

So , I tried to define u as arctg(1/√x) but I'm having trouble finding the du. The derivate of arctg(1/√x) is -1/2*x^(3/2), so it stays -1/2*x^(3/2)/1-(1/√x)^2 (converting to the derivate u'/1+u^2) ? I was thinking about getting the derivate of arctg(1/√x) and doing this by parts. dv = dx , u = arctg(1/√x)

2. Nov 24, 2012

### LCKurtz

The derivative of $\arctan(x^{-\frac 1 2})$ is$$\frac 1 {1+\frac 1 x}\left(-\frac 1 2 x^{-\frac 3 2}\right )= -\frac 1 2 \frac{x^{-\frac 1 2}}{x+1}$$That might help your integration by parts.

3. Nov 24, 2012

### ruiwp13

Thank you, now I got arctg(1/√x)*x - ∫((√x)/2x+2)*x . Do I need to make the integral by parts again?

4. Nov 24, 2012

### Ray Vickson

You should not get u'/1+u^2, which reads as
$$\frac{u'}{1} + u^2.$$ You should have gotten u'/(1+u^2), which reads as
$$\frac{u'}{1+u^2}.$$ Parentheses are important!

RGV

5. Nov 24, 2012

### ruiwp13

yes, I got this x.arctg(1/√x)+∫x/sqrt(x)*(2x+2) but now I'm having trouble on solving the integral :p any help?

6. Nov 24, 2012

### haruspex

Substitute for √x and use partial fractions.

7. Nov 25, 2012

### ruiwp13

I don't remeber how to use the partial fractions , it's that thing with A/(x-1) * B/(x+1) and A*(x+1)*B(x-1) thing? Can't remember quite well

8. Nov 25, 2012

### haruspex

It's that thing where you have one polynomial divided by another, P(x)/Q(x). You rewrite it as a sum of simpler fractions. http://en.wikipedia.org/wiki/Partial_fractions_in_integration

9. Nov 25, 2012

### ruiwp13

I transformed x/sqrt(x)*(2x+2) into sqrt(x)/2x+2 . Then I passed 2x+2 up and got Int (sqrt(x)*(2x+2)^-1) and multiplied and got Int (1/2sqrt(x)) + Int (sqrt(x)/2) but the second one is incorrect I think... Can't figure out why, tho.