Integration by Partial Fractions

  • #1

Homework Statement



[tex]\int_0^1 \! \frac{x-4}{x^2+3x+2} dx[/tex]


Homework Equations





The Attempt at a Solution



Factoring out the denominator...

[tex]x^2+3x+2 = (x+1)(x+2)[/tex]​


Breaking up the main fraction into the separate fractions. Since both are linear, only a single constant is needed on top.

[tex]\frac{A}{x+1}+\frac{B}{x+2}[/tex]​

Fining the coefficients (A and B) by choosing x values that make the other coefficient go to zero.

[tex]x-4 = A(x+2) + B(x+1)[/tex]

[tex]x=-2[/tex]
[tex]-6=B(-1)[/tex]
[tex]6=B[/tex]

[tex]x=-1[/tex]
[tex]-5=A(1)[/tex]
[tex]-5=A[/tex]​


Back in the integral they go. Also, breaking up the main integral into a sum of integrals.

[tex]\int \frac{-5}{x+1}dx + \int \frac{6}{x+2} dx[/tex]

[tex]-5ln(x+1) + 6ln(x+2) + c[/tex]


Now using the fundamental theorem of calculus to find the value.

[tex](-5ln(1+1) + 6ln(1+2))-(-5ln(0+1) + 6ln(0+2))[/tex]​


Result:

[tex]6ln(3)-11ln(2)[/tex]​

Wrong. Aw.
 

Answers and Replies

  • #2
jbunniii
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It looks OK to me. Who told you the answer was wrong?
 
  • #3
jbunniii
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P.S. Wolfram Alpha's answer agrees numerically with yours.
 
  • #4
The online homework system... though it gave me multiple options, and none of them were this. I suppose I could try to figure out which is equal, if my work is indeed correct.

http://mikederoche.com/temp/options.png [Broken]
 
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  • #5
Got it... equivalent to the second option.
 
  • #6
jbunniii
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It looks like they really want you to have a [itex]\ln(3/2)[/itex] term for some reason. OK then:

[tex]6 \ln(3) = 6 \ln(2(3/2)) = 6\ln(2) + 6\ln(3/2)[/tex]
 

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