Question

What is `f(x) = int 1/(x-3)-x/(x+4) dx` if `f(-1)=6`?

Answer 1

`f(x)=-x+lnabs(x-3)+4lnabs(x+4)+5-2ln2-4ln3`

Explanation:

`f(x)=lnabs(x-3)-int(y+4)/ydy + c = -y+4lnabsy+lnabs(x-3)+c=-x+4lnabs(x+4)+lnabs(x-3)+c`

Then

`f(-1)= 6` so

`6=1+4ln3+ln4+c`

`c=5-4ln3-3ln2`