Question

∫ sec^2 x tanx dx
-------------------
(1+sec^2 x) ^2

Answer 1

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

Answer 2

and not :

\[(\sec^2x - 1)^2?\]

Answer 3

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 ^_^

Answer 4

Correct me if I'm wrong :)