Integration by parts with limits? (verify my answer?)

limits [0,1]
Integral of 3x/(x+1)(x-2)

u=x^2 -x -2
(3+3du)/2 = 3x
QED new limits for U [0,3]

3/2 (of integral): 3du/u
3/2 [3logU] + C

.......here i am unsure if i just apply the U limits [0,3]. or replace U with my X-trinomial above & apply the X limits. Is my math to this point correct?

Integrate using partial fractions. For your choice of u, du = 2x - 1 dx which is not helpful.

3x/((x + 1)(x - 2)) = A/(x + 1) + B/(x - 2) = 1/(x + 1) + 2/(x - 2)

The integral is lnlx + 1l + 2lnlx - 2l = ln2 - 2ln2 = -ln2

You need to use partial fractions first to simply this into simpler functions which will be easier to integrate.

The fact that they give you a denominator already factored (i.e. (a+b)(a+c)) should be your clue to use partial fractions.

Here is the partial fraction break down of that function:

http://www.wolframalpha.com/input/?i=partial+fractions&a=*C.partial+fractions-_*Calculator.dflt-&f2=3x%2F%28%28x%2B1%29+%28x-2%29%29&x=8&y=6&f=ApartCalculator.rationalfunction_3x%2F%28%28x%2B1%29+%28x-2%29%29

Why would u wanna integrate this by parts?? note that 3x/(x+1)(x-2) = (1/x+1)+(2/x-2).
so the integral is ln(x+1) + 2ln(x-2).