Integration of 1+tanx/x+log(sec(x))?

1 Answer
Feb 12, 2018

The answer is log|x+log(secx)|

Explanation:

We perform this integration like this

int (1+tanx)/(x+log(secx))dx

Using substitution method :
1) Let x + log(secx) = t
2) On differentiating it with respect to x you get

1+(1/secx*secx.tanx)=dt/dx

This has become so because derivative of x w.r.t to x is 1 , derivative of log x is 1/x and chain rule is also used.

3) Simplify the above step by cancelling secx , and take dx to the LHS , you get

(1+tanx)dx = dt

Now (1+tanx)dx = dt and x + log(secx) = t
Substitute these into the original integral , you get ,

->intdt/t

On integration , intdt/t=log|t|

Substitute t as x+log(secx)
and your answer is ,

log|x+log(secx)|+c