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

1 Answer
Feb 3, 2016

f(x)=4/5lnAbs(x+2)-1/5lnAbs(x-3) -1-4/5ln(4)

Explanation:

Begin by splitting the fraction into partial fractions:

(x-2)/((x+2)(x-3)) = A/(x+2) + B/(x-3)

Now multiply by (x+2)(x+3) to find A and B.

x-2 = (x-3)A + (x+2)B

Set x=3 to find B.

(3)-2 = ((3)-3)A+ ((3)+2)B = 5B -> 5B =1
therefore B =1/5

Set x = -2 to find A so we will have:
-4 =-5A therefore A = 4/5

So we can now rewrite the integral in partial fractions and integrate:

int (x-2)/((x+2)(x-3))dx = int 4/(5(x+2)) + 1/(5(x-3))dx

therefore f(x)=4/5lnAbs(x+2)-1/5lnAbs(x-3) + C

Now you can use the condition specified: f(2) = -1

-> -1 = 4/5lnAbs(4) -1/5lnAbs(-1) + C

lnAbs(-1) = ln(1) = 0 which leaves us with:

C = -1-4/5ln(4) which means our final answer is:

f(x)=4/5lnAbs(x+2)-1/5lnAbs(x-3) -1-4/5ln(4)