Question

evaluate the integral

Answer 1

\[\int\limits_{?}^{?}(x ^{2} + 3x -44)dx/(x+3)(x+5)(3x-2)\]

Answer 2

partial fraction decomp it looks like; unless we can factor the top to cancel some bottoms

Answer 3

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

Answer 4

woohoo thats what I got. but now how do I find the actual values of the letters?

Answer 5

-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

Answer 6

when x = a number that zeros the other 2; use it and solve for that letter

Answer 7

but sometimes it leaves behind more than one variable

Answer 8

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)

Answer 9

sometimes it does, but use it first to get some solids :) with linears on the bottom, its not a problem

Answer 10

okay. for the last equation I got 4/9 - 42 = c(2)(2/3 + 5)

Answer 11

(4/9) +2 - 44 = C(11/3)(17/3)

(4/9) - 42 = C(11/3)(17/3)

Answer 12

-374/9 = C(187/9)

C = -374/189 reduce as wanted

Answer 13

wow how do you work so fast?

Answer 14

i use my head and mess up alot lol

Answer 15

lol oh...
do you think you can help me with another similar problem?

Answer 16

perhaps; but let me get thru this one first :)

Answer 17

okay! thanks for all of your help :)

Answer 18

we did C correctly right? that monster fraction scares me into thinking I messed it up lol

Answer 19

i typoed the 187 into a 189 ;)

Answer 20

C = 2 :)

Answer 21

-2

Answer 22

wait, in this problem.. there is no Dx+E?

Answer 23

so our integral becomes:

[S] 2/(x+3) - 1/(x+5) - 2/(3x-2) dx

Answer 24

the bottom is already factored for us into linear forms; linear form dont need the Dx+E form

Answer 25

ahh okay! Yayyyy. :D

Answer 26

if we had an irreducible quadratic in the denom; then we would use the Dx+E form above it to aid us

Answer 27

like.. (x-1)^3?

Answer 28

nope; that is a multiple of a linear and still only needs a A B C single atop it

Answer 29

really? how would it look.. A/x-1 + B/x-1 + c/x-1?

Answer 30

x^2 + 10 is irreducible quadratic; cant be factored to real linears

Answer 31

A/(x-1) + B/(x-1)^2 + C/(x-1)^3

Answer 32

ohh.. hmm . and that would be... Ax+B/x^2+10?

Answer 33

our integral has become:

2ln(x+3) - ln(x+5) -2ln(3x-2) +C

Answer 34

thats the final answer

Answer 35

thats the form we would use above it yes :)

Answer 36

we could probably turn this into one ln(yadayada)+C if we want, but this is good :)

Answer 37

wow thank you!!!!

Answer 38

youre welcome :)