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Evaluate the improper integral $$\int_0^\infty\frac{-38x}{(2x^2+9)(3x^2+4)} dx $$

I thought about doing this through partial fractions decomposition. However, when I tried, I got some really awful numbers like $\dfrac{114}{19}$. Did I make a mistake somewhere in my math, or is there an easier method to solving this? Also, this is a multiple choice question and the answers all involve $\ln$. Is that another indicator that the best method is partial fractions?

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    $\begingroup$ Substitute $u=x^2$ and then the partial fractions is easy. $\endgroup$ – Ted Shifrin Apr 13 '14 at 17:48
  • $\begingroup$ @TedShifrin but then, wouldn't that leave me with a square root u in the numerator? $\endgroup$ – Chrysanthemum Apr 13 '14 at 17:49
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    $\begingroup$ No, the derivative is gonna save you @Chrysanthemum $\endgroup$ – J.R. Apr 13 '14 at 17:49
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    $\begingroup$ What's so terrible about $\frac{114}{19}$? $\endgroup$ – Javier Apr 13 '14 at 17:52
  • $\begingroup$ @JavierBadia Write the digit three times, one after the other, and you have a number that some people consider evil. $\endgroup$ – Daniel Fischer Apr 13 '14 at 17:54
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Substituting $u=x^2$, $$I=\int_0^\infty \frac{-38x}{(2x^2+9)(3x^2+4)} dx = \int_0^\infty \frac{-19}{(2u+9)(3u+4)} du$$

Now,

$$\frac{-19}{(2u+9)(3u+4)}=\frac{2}{2u+9}-\frac{3}{3u+4}$$

Thus

$$I=\lim_{L\rightarrow\infty}\left[ \log(2u+9) - \log(3u+4)\right|_0^L=\left(\lim_{u\rightarrow\infty} \log(2u+9)-\log(3u+4) \right) -2\log(3/2)$$

Further

$$\lim_{u\rightarrow\infty} \log(2u+9)-\log(3u+4)=\lim_{u\rightarrow\infty} \log\frac{2u+9}{3u+4}=\log\frac{2}{3}=-\log\frac{3}{2}$$

Therefore,

$$I=-3\log\frac{3}{2}$$

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  • $\begingroup$ you should replace u with L in the limit $\endgroup$ – Dylan Apr 13 '14 at 18:50
  • $\begingroup$ Yes typo, thank you. @DylanDang $\endgroup$ – J.R. Apr 14 '14 at 6:57

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