Request for crazy integrals I'm a sucker for exotic integrals like the one evaluated in this post. I don't really know why, but I just can't get enough of the amazing closed forms that some are able to come up with.
So, what are your favorite exotic integral identities, and how do you prove them?

Here are some links to a few integrals: 1
(Big list, but not all of them got the right answer).
From AoPS: 2, 3 , 4. Some that are solvable with Feynman's trick: here.
As for my favourites (most of them appeared on Romanian Mathematical Magazine), some are: $$I_1=\int_0^\frac{\pi}{2} \frac{\arctan(\tan x\sec x)}{\tan x +\sec x}dx=\frac{\pi}{2}\ln 2 -\frac{\pi}{6}\ln(2+\sqrt 3)$$
$$I_2=\int_0^\infty \exp\left(-\frac{3x^2+15}{2x^2+18}\right)\cos\left(\frac{2x}{x^2+9}\right)\frac{dx}{x^2+1}=\frac{\pi}{e}$$
$$I_3=\int_0^1 \frac{\ln^2 (1+x) (\ln^2 (1+x) +6\ln^2(1-x))}{x}dx=\frac{21}{4}\zeta(5)$$
$$I_4=\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2}\frac{dx}{\sqrt x}=-\frac{\pi}{24}$$
$$I_5=\int_0^\infty \frac{1-\cos x}{8-4x\sin x +x^2(1-\cos x)}dx=\frac{\pi}{4}$$
$$I_6=\int_0^\infty \frac{\arctan x}{x^4+x^2+1}dx=\frac{\pi^2}{8\sqrt{3}}-\frac{2}{3}G+\frac{\pi}{12}\ln(2+\sqrt{3})$$
$$I_7=\int_0^\infty \frac{\ln(1+x)}{x^4-x^2+1}dx=\frac{\pi}{6}\ln(2+\sqrt 3)+\frac23 G -\frac{\pi^2}{12 \sqrt 3}$$
$$I_8=\int_0^1 \frac{\ln(1-x^2)\ln(1+x^2)}{1+x^2}dx=\frac{\pi^3}{32}-3G\ln 2+\frac{\pi}{2}\ln^22.$$
$$I_{9}=\int_0^{\frac{\pi}{4}} \ln\left(2+\sqrt{1-\tan^2 x}\right)dx = \frac{\pi}{2}\ln\left(1+\sqrt{2}\right)+\frac{7\pi}{24}\ln2-\frac{\pi}{3}\ln\left(1+\sqrt{3}\right)-\frac{G}{6}$$
$$I_{10}=\int_{-\infty}^\infty \frac{\sin \left(x-\frac{1}{x}\right) }{x+\frac{1}{x}}dx=\frac{\pi}{e^2}$$
$$I_{11}=\int_{-\infty}^\infty \frac{\cos \left(x-\frac{1}{x}\right) }{\left(x+\frac{1}{x}\right)^2}dx=\frac{\pi}{2e^2}$$
$$I_{12}=\int_0^1 \frac{\ln(1-x)\ln(1-x^4)}{x}dx=\frac{67}{32}\zeta(3)-\frac{\pi}{2} G$$
$$I_{13}=\int_0^\frac{\pi}{2} x^2 \sqrt{\tan x}dx=\frac{\sqrt{2}\pi(5\pi^2+12\pi\ln 2 - 12\ln^22)}{96}$$
$$I_{14}=\int_0^\frac{\pi}{4} \operatorname{arcsinh} (\sin x) dx=G-\frac58\operatorname{Cl}_2\left(\frac{\pi}{3}\right)$$
$$I_{15}=\int_0^\frac{\pi}{2} x \arcsin \left(\sin x-\cos x\right)dx=\frac{\pi^3}{96}+\frac{\pi}{8}\ln^2 2$$
$$I_{16}=\int_0^\infty \int_0^\infty \frac{\ln(1+x+y)}{xy\left((1+x+y)(1+1/x+1/y)-1\right)}dxdy=\frac72 \zeta(3)$$
Where $G$ is Catalan's constant and $\operatorname{Cl}_2 (x)$ is the Clausen function.
