# Does this make sense

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?

$$I = \int {\frac{6}{{\left( {x^2 + 3} \right)^2 }}dx}$$
$$x = \sqrt 3 \tan \theta \Rightarrow dx = \sqrt 3 \sec ^2 \theta d\theta$$