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?
Best Answer
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,
$$I=-3\log\frac{3}{2}$$