Calc 2 Integration Problem


by Tilted
Tags: calc, integration
Tilted
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#1
Mar7-12, 08:56 PM
P: 6
1. The problem statement, all variables and given/known data
[itex]\int[/itex][itex](x^2-2x+3)dx/(x^3-x^2-x-2)[/itex]


2. Relevant equations
Trig substitution/ partial fractions?



3. The attempt at a solution

I used partial fractions to reduce the integral down to:

[itex]\int[/itex][itex]dx/(x-2)[/itex]+[itex]\int[/itex][itex](x-1)dx/(x^2+x+1)[/itex]

The first integral is easy enough, but the second one I'm not sure where to start.

Thanks in advance!
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Dick
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#2
Mar7-12, 09:27 PM
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I don't agree with all of your numbers there but the general form is right. For the quadratic part you want to complete the square in the denominator, do substitution for the squared part, then split it up and use a trig substitution.
jzachey
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#3
Mar8-12, 07:42 AM
P: 4
using partial fractions I obtained
(x^2-2x+3)
-----------dx
((x-2)(x^2+x+1)

A bx+c
-- + --------
x-2 x^2+x+1

A=3/7
B=4/7
c=-9/7
this is my attempt, i how it helps somwhat, I'm not sure if it's correct so I apologize early on.

SammyS
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#4
Mar8-12, 09:45 AM
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Calc 2 Integration Problem


Quote Quote by jzachey View Post
using partial fractions I obtained
(x^2-2x+3)
-----------dx
((x-2)(x^2+x+1)

A Bx+C
-- + --------
x-2 x^2+x+1
A                 Bx+C
--       +     --------
x-2           x^2+x+1
A=3/7
B=4/7
C=-9/7
this is my attempt, i how it helps somwhat, I'm not sure if it's correct so I apologize early on.
Those values for A, B, and C, are correct.
jzachey
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#5
Mar8-12, 05:40 PM
P: 4
Quote Quote by SammyS View Post
A                 Bx+C
--       +     --------
x-2           x^2+x+1
Those values for A, B, and C, are correct.
Oh thank you so this means
∫[itex]\frac{3}{7(x-2)}[/itex]+[itex]\frac{4x-9}{7(x+x+1)}[/itex]
From here you integrate
SammyS
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#6
Mar8-12, 06:04 PM
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Quote Quote by jzachey View Post
Oh thank you so this means
[itex]\displaystyle \int\left(\frac{3}{7(x-2)}+\frac{4x-9}{7(x^2+x+1)}\right)dx[/itex]
From here you integrate
There's a typo in your post, corrected (and reformatted) above.


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