What is #f(x) = int 1/(x-3)-x/(x+4) dx# if #f(-1)=6 #? Calculus Techniques of Integration Evaluating the Constant of Integration 1 Answer Alberto P. Nov 9, 2016 #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# Answer link Related questions How do you find the constant of integration for #intf'(x)dx# if #f(2)=1#? What is a line integral? What is #f(x) = int x^3-x# if #f(2)=4 #? What is #f(x) = int x^2+x-3# if #f(2)=3 #? What is #f(x) = int xe^x# if #f(2)=3 #? What is #f(x) = int x - 3 # if #f(2)=3 #? What is #f(x) = int x^2 - 3x # if #f(2)=1 #? What is #f(x) = int 1/x # if #f(2)=1 #? What is #f(x) = int 1/(x+3) # if #f(2)=1 #? What is #f(x) = int 1/(x^2+3) # if #f(2)=1 #? See all questions in Evaluating the Constant of Integration Impact of this question 1322 views around the world You can reuse this answer Creative Commons License