Question

solved the integration and give me result

Answer 1

if I = e^ax. { af(x) + f'(x) }.dx then,
I = e^ax.f(x) + C
use this concept here,

I = e^x { tanx + lnsecx }.dx

Let lnsecx = f(x)
differentiate with respect to x
1/secx {secx.tanx } = f'(x)
f'(x) = tanx

hence integration convert in the form of
I = e^x { f(x) + f'(x) }.dx

hence,
I = e^f(x) + C
but f(x) = lnsecx put it .

I = e^x.lnsecx + C

Answer 2

Answer:

Hello

Step-by-step explanation:

Hi..