sorry, but i dont integrate with substituting. I learnt another method that is... let me give an example
i add the power by 1 and divide the whole function by the new power, and divide it once again by the derivative.
however, this function has x in both the denominator and the numerator, which is what makes it tough for me
You have been shown how to do this question: The integrand can be written as (see below) and you claim in an earlier post to be able to integrate expressions like these. So there shouldn't be any problem.