Really advanced techniques of integration (definite or indefinite) Okay, so everyone knows the usual methods of solving integrals, namely u-substitution, integration by parts, partial fractions, trig substitutions, and reduction formulas. But what else is there? Every time I search for "Advanced Techniques of Symbolic Integration" or "Super Advanced Integration Techniques", I get the same results which end up only talking about the methods mentioned above. Are there any super obscure and interesting techniques for solving integrals?
As an example of something that might be obscure, the formula for "general integration by parts " for $n$ functions $f_j, \ j = 1,\cdots,n$ is given by
$$
\int{f_1'(x)\prod_{j=2}^n{f_j(x)}dx} = \prod_{i=1}^n{f_i(x)} - \sum_{i=2}^n{\int{f_i'(x)\prod_{\substack{j=1 \\ j \neq i}}^n{f_j(x)}dx}}
$$
which is not necessarily useful nor difficult to derive, but is interesting nonetheless.
So out of curiosity, are there any crazy unknown symbolic integration techniques?

The use of residues to evaluate real integrals. There are many theorems that can be applied to various cases. If I get to it, I'll post some examples from this website. Later.
OK, here is an example. First a theorem. 
If $p(x) \ \rm{and} \ q(x) $ are real polynomials such that the degree of $q(x)$ is at least 2 more than the degree of $p(x),$ and if $q(x)=0$ has no real roots, then 
$$ \int_{- \infty}^{\infty} \frac{p(x)}{q(x)}dx  = 2 \pi i \sum{ \rm{residues \ of \ } }p(z)/q(z) \ \rm{at \ its \ poles \ in \ the \ upper \ half \ plane.}$$
Now apply this to find $$\int_{- \infty}^{\infty} \frac{x^2}{\left(x^2+a^2 \right) \left(x^2+b^2 \right) } dx$$
The only poles of $$\frac{z^2}{\left(z^2+a^2 \right) \left(z^2+b^2 \right) } $$
are at $z= \pm ai, z= \pm bi.$ Only $ai$ and $bi$ are in the upper half plane. For $z=ai$ the residue is
$$ \lim\limits_{z \to ai} (z-ai) \frac{z^2}{\left(z-ai \right)\left(z+ai \right) \left(z^2+b^2 \right)}=\frac{-a^2}{2ai \left(-a^2+b^2 \right) }=\frac{a}{2i \left(a^2-b^2 \right)}$$
Similarly, the other residue is $$ \frac{b}{2i \left(b^2-a^2 \right)}$$
Therefore, the value of the integral is
$$ 2 \pi i \left[ \frac{a}{2i \left(a^2-b^2 \right)  } + \frac{b}{2i \left(b^2-a^2 \right)} \right] =\frac{\pi}{a+b}$$
$$ \ $$
Reference: $ Advanced \ Engineering \ Mathematics, \rm{Wylie \ and \ Barrett.}$ 
