Compute the following integral.

Using the previous exercise we know that with the substitution we obtain

This time, since we make the substitution

This gives us, for ,

Using partial fractions we can rewrite the integrand in this to obtain,

i guess denominator in log is wrong it should be -1-acosx and the sign is to be positive at the end where it was changed to negative for perfect answer

There are 2 typos. The $s$ in the third row should be $u$, and there should be no minus sign at the binning of the last 2 rows (i.e. |x| = |-x|)