What is #f(x) = int 1/(x+3)-1/(x-2) dx# if #f(-1)=3 #? Calculus Techniques of Integration Evaluating the Constant of Integration 1 Answer Ananda Dasgupta Apr 3, 2018 #f(x)=ln|3/2 (x+3)/(x-2)|+3# Explanation: #f(x) = int [1/(x+3)-1/(x-2)] dx = ln|x+3|-ln|x-2|+C# #qquad = ln|(x+3)/(x-2)|+C# Since #f(-1) = 3#, we have #3 = ln|(-1+3)/(-1-2)|+C = ln|-2/3|+C implies C = 3-ln|2/3|# Thus #f(x) = ln|(x+3)/(x-2)|+3-ln|2/3|=ln|3/2 (x+3)/(x-2)|+3# 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 1315 views around the world You can reuse this answer Creative Commons License