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

Answer 1

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

#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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Answer 2

Emilia Aguilar

The function ( f(x) = \int \frac{1}{x+3} - \frac{1}{x-4} , dx ) with ( f(-2) = 3 ) can be evaluated as ( f(x) = \ln{|x+3|} - \ln{|x-4|} + C ), where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.