What is #f(x) = int 1/(x+3)-1/(x-4) dx# if #f(-2)=3 #?

1 Answer
Trevor Ryan.
Jan 9, 2016

#therefore f(x)=ln|x+3|-ln|x-4|+3-ln6#.

Explanation:

#f(x)=int(1/(x+3)-1/(x-4))dx=int1/(x+3)dx-int1/(x-4)dx#

#=ln|x+3|-ln|x-4|+C#.

But since #f(-2)=3# is given as an initial/boundary condition, we can substitute it into the integral to find the constant of integration :

#therefore ln|-2+3|+ln|-2-4|+C=3#

#therefore0+ln6+C=3,=>C=3-ln6#.

#therefore f(x)=ln|x+3|-ln|x-4|+3-ln6#.

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