evaluate the integral

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evaluate the integral

Mathematics
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\[\int\limits_{?}^{?}(x ^{2} + 3x -44)dx/(x+3)(x+5)(3x-2)\]
partial fraction decomp it looks like; unless we can factor the top to cancel some bottoms
x^2 +3x -44 = A(x+5)(3x-2) ; x = -3 x^2 +3x -44 = B(x+3)(3x-2) ; x = -5 x^2 +3x -44 = C(x+3)(x+5) ; x = 2/3

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woohoo thats what I got. but now how do I find the actual values of the letters?
-44 = A(2)(-11) -44/-22=A A = 2 ---------------------- 25 -59 = B(-2)(-17) -34 = B(34) B = -1 x^2 +3x -44 = C(x+3)(x+5) ; x = 2/3
when x = a number that zeros the other 2; use it and solve for that letter
but sometimes it leaves behind more than one variable
x^2 +3x -44 = C(x+3)(x+5) ; x = 2/3 (2/3)^2 +3(2/3) - 44 = C(2/3 +3)(2/3 +5)
sometimes it does, but use it first to get some solids :) with linears on the bottom, its not a problem
okay. for the last equation I got 4/9 - 42 = c(2)(2/3 + 5)
(4/9) +2 - 44 = C(11/3)(17/3) (4/9) - 42 = C(11/3)(17/3)
-374/9 = C(187/9) C = -374/189 reduce as wanted
wow how do you work so fast?
i use my head and mess up alot lol
lol oh... do you think you can help me with another similar problem?
perhaps; but let me get thru this one first :)
okay! thanks for all of your help :)
we did C correctly right? that monster fraction scares me into thinking I messed it up lol
i typoed the 187 into a 189 ;)
C = 2 :)
-2
wait, in this problem.. there is no Dx+E?
so our integral becomes: [S] 2/(x+3) - 1/(x+5) - 2/(3x-2) dx
the bottom is already factored for us into linear forms; linear form dont need the Dx+E form
ahh okay! Yayyyy. :D
if we had an irreducible quadratic in the denom; then we would use the Dx+E form above it to aid us
like.. (x-1)^3?
nope; that is a multiple of a linear and still only needs a A B C single atop it
really? how would it look.. A/x-1 + B/x-1 + c/x-1?
x^2 + 10 is irreducible quadratic; cant be factored to real linears
A/(x-1) + B/(x-1)^2 + C/(x-1)^3
ohh.. hmm . and that would be... Ax+B/x^2+10?
our integral has become: 2ln(x+3) - ln(x+5) -2ln(3x-2) +C
thats the final answer
thats the form we would use above it yes :)
we could probably turn this into one ln(yadayada)+C if we want, but this is good :)
wow thank you!!!!
youre welcome :)

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