# Does this make sense

1. Sep 26, 2005

### badtwistoffate

in int[6/(x^2+3)^2] i use 3tanU=x, dx=3sec(u)^2 du
using that i get to .... int[1+cos2u]= u + sin2u/2 = u +sin u cos u... but subing back in with trig doesnt give me the right answer... any help with this method?

2. Sep 26, 2005

### Benny

$$I = \int {\frac{6}{{\left( {x^2 + 3} \right)^2 }}dx}$$

Instead of setting x = 3tan(u) try...

$$x = \sqrt 3 \tan \theta \Rightarrow dx = \sqrt 3 \sec ^2 \theta d\theta$$

3. Sep 26, 2005

### badtwistoffate

darn it, after looking at my work i was afraid someone would say that... as i didnt relize that would be better until i finished it and noticed it was coming out right.... does that mean you cant do it my way:-/

4. Sep 26, 2005

### Benny

There's really no reason why it can't be done your way. It's just that some substitutions will require you to evaluate much more complicated integrals than if you were to use a substitution which obviously simplifies the integrand.

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