Integral (x^2 + 7x + 12)/(x + 4)

  • Thread starter b0rsuk
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  • #1
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Homework Statement



Hello. I have a problem with a innocent-looking integral:

[tex] \int\frac{x^2 + 7x + 12}{x + 4} dx [/tex]

It doesn't look like i can use the law of sines, because the numerator is of higher order than the denominator. It doesn't look like the numerator is a multiple of the denominator or vice versa. It doesn't look like I can split the relatively complex numerator to an useful product.

It looks like it relies on some high school level math for solving polynomials / fractions, but my math is rather rusty in some places and I can't seem to recall anything. I could look for the method if I knew its name. Sometimes it's best to ask a human being.

Homework Equations



It's from introductory examples, which means I'm not yet allowed to use:
- Integration by substitution
- Integration by parts


The Attempt at a Solution



[tex]\int\frac{x^2 + 7x + 12}{x + 4} dx = \int \frac{x^2}{x + 4} + \frac{7x}{x + 4} + \frac{12}{x + 4} dx =[/tex]
[tex]=\int \frac{x^2}{x + 4} + \frac{7x}{x + 4}dx + 12 \int \frac{dx}{x + 4} =[/tex]
[tex]= \int \frac{x^2}{x + 4} + \frac{7x}{x + 4}dx + 12 \ln |x + 4| + C_1[/tex]

No, I can't get very far.
 

Answers and Replies

  • #2
353
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The top can be factored, so that the denominator is canceled.
 
  • #3
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Oh, I gave up too soon.
It's not too hard if you start with the assumption that (x + 4) is a factor in numerator. Case closed :-)
 
  • #4
Dick
Science Advisor
Homework Helper
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Oh, I gave up too soon.
It's not too hard if you start with the assumption that (x + 4) is a factor in numerator. Case closed :-)

Even if (x+4) were not a factor of the numerator, you can still do it by dividing the numerator by the denominator giving you a quotient and remainder.
 

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