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Find the integral of x/(x^2 +4x +13)

  • #1
1. Find the integral of x/(x^2 +4x +13)

I'm not sure if I need to complete the square or what. I really don't know where to begin.
 

Answers and Replies

  • #2
eumyang
Homework Helper
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10


Try rewriting the integrand as a difference of two fractions.
 
  • #3
HallsofIvy
Science Advisor
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Yes, start by completing the square. That will reduce it to something like [itex](x- a)^2+ b[/itex]. Then let u= x- a so that denominator becomes [itex]u^2+ b[/itex] and the numerator becomes x= u+ a:
[tex]\int \frac{x}{x^2+ 4x+ 13}= \int \frac{u}{u^2+ b}du+ \int\frac{a}{u^2+ b}du[/tex]
The first integral can be done by the substitution [itex]v= u^2+ b[/itex] and the second is an arctangent.
 
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  • #4
hunt_mat
Homework Helper
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First however note that:
[tex]
\frac{x}{x^{2}+4x+13}=\frac{1}{2}\frac{2x+4}{x^{2}+4x+13}-\frac{2}{x^{2}+4x+13}
[/tex]
When integrating this expression, the first term is of the form f'(x)/f(x) which integrates to log(f(x)). Complte the square for the second term.
 

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