Asked by Ace
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.
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.