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How would I integrate $$\int \frac{x}{x^2-10x+50} \, dx$$ I am not sure on how to start the problem

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  • $\begingroup$ Partial fractions! $\endgroup$ – Laars Helenius Mar 9 '15 at 17:06
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    $\begingroup$ @LaarsHelenius the quadratic has two complex roots, so unless complex integration is allowed I don't think that would work $\endgroup$ – graydad Mar 9 '15 at 17:11
  • $\begingroup$ Ah, I see. I thought I saw $c=21$ earlier. So either I'm an idiot or the question was edited after my comment. For the record, I'm not willing discount either option! Lol $\endgroup$ – Laars Helenius Mar 9 '15 at 17:41
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I would complete the square in the denominator first.

$$ \displaystyle \int \dfrac{x}{x^2-10x+25+25}dx = \int \dfrac{x}{(x-5)^2+25}dx $$

Let $u=x-5$,

$$ \displaystyle \int \dfrac{u+5}{u^2+25}du = \int \dfrac{u}{u^2+25}du + \int \dfrac{5}{u^2+25}du$$

the first integral you can solve by doing one more substitution, the second is just arctangent.

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write the integral as $$\int\frac{x}{x^2 - 10x +50}dx = \int \left(\frac{2x-10}{2(x^2-10x+50)} + \frac{5}{x^2-10x+50}\right)dx$$ Let $u = x^2 -10x +50, \; \text{then } du = 2x-10$ $$\frac{1}{2}\int\frac{1}{u}du + 5\int \frac{1}{(x-5)^2 + 25}dx$$ $$\frac{\log (u)}{2}+5\int\frac{1}{\frac{(x-5)^2}{25}+1}dx$$ $$\frac{1}{2}\log(x^2-10x+50) + \arctan\left(\frac{x-5}{5}\right) + C$$

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  • $\begingroup$ I think you want 5 in the 2nd numerator (instead of x), but this is a good method to solve this. $\endgroup$ – user84413 Mar 9 '15 at 23:21
  • $\begingroup$ yeah, you are right $\endgroup$ – Alexander Mar 10 '15 at 0:32
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Complete the square in the denominator:

$$x^2 - 10x + 50 = x^2 -10 x + 25 + 25 = (x-5)^2 + 5^2$$

Put $x-5 = 5\tan\theta\implies dx = 5\sec^2 x$.

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