Question

help?
Evaluate the integral (shown below)?

Answer 1

\[\int\limits \frac{ -45x^2-27x+18 }{ x^3+2x^2 }\]

Answer 2

i got this but its wrong? 18*log(x)+(-2/x)+2*x-27*log(x)+C

Answer 3

use partial fraction to simplify that expresson

Answer 4

how do i do that?

Answer 5

\[\frac{-45x^2-27x+18}{x^3+2x^2}=\frac{-3(15x-6)(x+1)}{x^2(x+2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+2}\\
-3(15x-6)(x+1)=A(x)(x+2)+B(x+2)+C(x^2)\]
Solve for the constants, then integrate term by term.

Answer 6

Note: I only factored the numerator to see if there would be any removable factors in both numerator and denominator. Since there were none, you can leave the numerator un-factored. However, keeping it the way I wrote it may be helpful depending on the method you use to solve for A, B, and C.

Answer 7

Bx+C at the second

Answer 8

@experimentX, we've got a repeated root in the denominator...
http://www.purplemath.com/modules/partfrac2.htm

Answer 9

ya i did that and then got b=2 and a=18 and c=-108/4?

Answer 10

i thought we would add up linear term in repeated roots too ..

Answer 11

im confused?

Answer 12

I get A = -3, B = -9, and C = -27.

Answer 13

@experimentX I think that's just the case for irreducible quadratic factors. According to that link it looks like repeated factors of irreducible quadratics are treated the same way as linear factors.

Answer 14

\[\int\limits -3/x+-9/x^2+-27?x+2 dx? \]

Answer 15

But I guess it's taken into account with the first two terms, since
\[\frac{Ax+B}{x^2}=\frac{A}{x}+\frac{B}{x^2}\]

Answer 16

maybe I misremembered something ... i don't do PF enough since I am addicted to wolf

http://www.wolframalpha.com/input/?i=partial+fractions+%28%E2%88%9245x%5E2%E2%88%9227x%2B18%29%2F%28x%5E3%2B2x%5E2%29

Answer 17

@mags093, do you know how to get the same A, B, and C that I did?

Answer 18

no?

Answer 19

Ah, looks like I got it wrong too. Just a sec...

Answer 20

ok :)

Answer 21

@SithsAndGiggles i guess it is.

Answer 22

Okay, I made some algebraic mistake the last time, so I'll use the more dependable method:
\[-45x^2-27x+18=A(x)(x+2)+B(x+2)+C(x^2)\\
-45x^2-27x+18=Ax^2+2Ax+Bx+2B+Cx^2\\
-45x^2-27x+18=(A+C)x^2+(2A+B)x+2B\]
Matching up the coefficients, you get
\[\begin{cases}A+C=-45\\ 2A+B=-27\\ 2B=18\end{cases}\]
Solving the systems gives you A = -18, B = 9, and C = -27.

So, the integral is rewritten as
\[\begin{align*}\int\frac{-45x^2-27x+18}{x^3+2x}dx&=\int\frac{-18}{x}dx+\int\frac{9}{x^2}dx+\int\frac{-27}{x+2}dx\\
&=-18\int\frac{dx}{x}+9\int\frac{dx}{x^2}-27\int\frac{dx}{x+2}\end{align*}\]
The remaining integrals are simple

Answer 23

i cant work that out? it keeps telling me im wrong?

Answer 24

-9(1/x+2log(x) +3log(x+2)) + C