## anonymous 5 years ago ∫ sec^2 x tanx dx ------------------- (1+sec^2 x) ^2

1. anonymous

are you sure it's over $(1+\sec^2x)^2$?

2. anonymous

and not : $(\sec^2x - 1)^2?$

3. anonymous

if not, then you can do it this way ^_^: 1) substitute sec^2x in the denominator with (1+tan^2x) and you'll get : $\int\limits_{}^{}(\sec^2xtanx)/(1+1+\tan^2x)^2dx$ $=\int\limits_{}^{}(\sec^2xtanx)/(2+\tan^2x)^2dx$ 2) let u = tanx and du = sex^2x dx: $=\int\limits_{} (u)/(2+u^2)^2 du$ get (2+u^2)^2 up and you'll have : $=\int\limits_{} u(2+u^2)^{-2}du$ now you can use integration by parts by letting the following as given : - f' = (2+u^2)^(-2) and g = u find f and g' then use the general equation which is: $\int\limits_{}f'gdx = fg - \int\limits_{}fg'dx$ or partial fractions to solve this integration :) I hope I have given you hint, you can take it from here ^_^

4. anonymous

Correct me if I'm wrong :)