<s>[INST] How do I evaluate the integral $\int_0^{\infty}\frac{x^5\sin(x)}{(1+x^2)^3}dx$? I have no idea how to start, it looks like integration by parts won't work.
$$\int_0^{\infty}\frac{x^5\sin(x)}{(1+x^2)^3}dx$$
If someone could shed some light on this I'd be very thankful.
 [/INST] I try to tackle the problem by differentiation under integral sign.
Using the famous result:
$$
\int_0^{\infty} \frac{\cos a x}{x^2+t^2} dx=\frac{\pi}{2 t} e^{-a t}
$$
where $a,t>0.$
we get the result of the integral
$$
\int_0^{\infty} \frac{\cos a x}{1+t x^2} d x=\frac{\pi}{2} \frac{1}{\sqrt{t} e^{\frac{a}{\sqrt t}}}
$$
Differentiating both sides w.r.t $t$ twice at $t=1$ yields
$$
\int_0^{\infty} \frac{x^4 \cos (a x)}{\left(1+x^2\right)^3} d x=\frac{\pi\left(3-5 a+a^2\right)}{16 e^a}
$$
Differentiating both sides w.r.t $a$ once yields
$$
\frac{\partial}{\partial a} \int_0^{\infty} \frac{x^4 \cos (a x)}{\left(1+x^2\right)^3} d x =\frac{\pi}{16} \frac{\partial}{\partial a}\left(\frac{3-5 a+a^2}{e^a}\right)
$$
Putting $a=1$ yields
$$
-\int_0^{\infty} \frac{x^5 \sin x}{\left(1+x^2\right)^3}=\frac{\pi}{16}\left(-\frac{2}{e}\right)
$$
Hence
$$\boxed{\int_0^{\infty} \frac{x^5 \sin x}{\left(1+x^2\right)^3}=\frac{\pi}{8e}}$$
</s>