Posted by **Ace** on Wednesday, June 13, 2007 at 1:06am.

How do I solve

Integral of 7/(16-x^2)

I know I must break down (16-x^2) into (x+4)(-x+4), but after I do that what is next?

Using the method of partial fractions, convince yourself that

7/(16-x^2)= (7/8)*[1/(4+x) + 1/(4-x)]

The two additive

terms can be integrated by the method of substitution.

4+x -> u etc. You will end up with terms that are the log of 4+x and 4-x.