What is #f(x) = int 1/(x+3)-1/(x-2) dx# if #f(-1)=3 #?

1 Answer
Apr 3, 2018

Answer:

#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#