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How to find integrals like int (1-x)(x^2-4)dx

  • Thread starter amcavoy
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  • #1
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I already know the answer to this, but would like your opinion on the quickest way to calculate integrals like these. I am finding myself evaluating similar integrals in my diffeq homework, and the method that I used (parts and partial fractions) is a mess and takes way too long (if it were to appear on a test).

[tex]\int\left(\frac{1-x}{x^2-4}\right)dx[/tex]

Thanks.
 

Answers and Replies

  • #2
LeonhardEuler
Gold Member
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In general, partial fractions is the only way to go. In this particular case, you can split the integral up like this to do it pretty fast:
[tex]\int-\frac{x-1}{x^2-4}dx[/tex]
[tex]=-\int\frac{x-2}{(x+2)(x-2)}+\frac{2}{2(x+2)(x-2)}dx[/tex]
[tex]=-\int\frac{1}{x+2}+\frac{2+x}{2(x+2)(x-2)}-\frac{x}{2(x^2-4)}}dx[/tex]
[tex]=-\int\frac{1}{x+2}+\frac{1}{2(x-2)}-\frac{x}{2(x^2-4)}}dx[/tex]
Now you can directly compute the first two parts as logs and the last as a log by recognizing x as one quarter the derivative of the denomenator
 
  • #3
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LeonhardEuler said:
In general, partial fractions is the only way to go. In this particular case, you can split the integral up like this to do it pretty fast:
[tex]\int-\frac{x-1}{x^2-4}dx[/tex]
[tex]=-\int\frac{x-2}{(x+2)(x-2)}+\frac{2}{2(x+2)(x-2)}dx[/tex]
[tex]=-\int\frac{1}{x+2}+\frac{2+x}{2(x+2)(x-2)}-\frac{x}{2(x^2-4)}}dx[/tex]
[tex]=-\int\frac{1}{x+2}+\frac{1}{2(x-2)}-\frac{x}{2(x^2-4)}}dx[/tex]
Now you can directly compute the first two parts as logs and the last as a log by recognizing x as one quarter the derivative of the denomenator
Well it looks like the only way to save time here is to use my TI-89 :smile:.

Thanks for the reply Euler.
 

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