Lesser-known integration tricks I am currently studying for the GRE math subject test, which heavily tests calculus. I've reviewed most of the basic calculus techniques (integration by parts, trig substitutions, etc.) I am now looking for a list or reference for some lesser-known tricks or clever substitutions that are useful in integration. For example, I learned of this trick 
$$\int_a^b f(x) \, dx = \int_a^b f(a + b -x) \, dx$$
in the question Showing that $\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx$, when $f$ is even 
I am especially interested in tricks that can be used without an excessive amount of computation, as I believe (or hope?) that these will be what  is useful for the GRE.

When integrating rational functions by partial fractions decomposition, the trickiest type of antiderivative that one might need to compute is
$$I_n = \int \frac{dx}{(1+x^2)^n}.$$
(Integrals involving more general quadratic factors can be reduced to such integrals, plus integrals of the much easier type $\int \frac{x \, dx}{(1+x^2)^n}$, with the help of substitutions of the form $x \mapsto x+a$ and $x \mapsto ax$.)
For $n=1$, we know that $I_1 = \int \frac{dx}{1+x^2} = \arctan x + C$, and the usual suggestion for finding $I_n$ for $n \ge 2$ is to work one's way down to $I_1$ using the reduction formula
$$
I_n = \frac{1}{2(n-1)} \left( \frac{x}{(1+x^2)^{n-1}} + (2n-3) \, I_{n-1} \right)
.
$$
However, this formula is not easy to remember, and the computations become quite tedious, so the lesser-known trick that I will describe here is (in my opinion) a much simpler way.
From now on, I will use the abbreviation
$$T=1+x^2.$$
First we compute
$$
\frac{d}{dx} \left( x \cdot \frac{1}{T^n} \right)
= 1 \cdot \frac{1}{T^n} + x \cdot \frac{-n}{T^{n+1}} \cdot 2x
= \frac{1}{T^n} - \frac{2n x^2}{T^{n+1}}
= \frac{1}{T^n} - \frac{2n (T-1)}{T^{n+1}}
\\
= \frac{1}{T^n} - \frac{2n T}{T^{n+1}} + \frac{2n}{T^{n+1}}
= \frac{2n}{T^{n+1}} - \frac{2n-1}{T^n}
.
$$
Let us record this result for future use, in the form of an integral:
$$
\int \left( \frac{2n}{T^{n+1}} - \frac{2n-1}{T^n} \right) dx = \frac{x}{T^n} + C
.
$$
That is, we have
$$
\begin{align}
\int \left( \frac{2}{T^2} - \frac{1}{T^1} \right) dx &= \frac{x}{T} + C
,\\
\int \left( \frac{4}{T^3} - \frac{3}{T^2} \right) dx &= \frac{x}{T^2} + C
,\\
\int \left( \frac{6}{T^4} - \frac{5}{T^3} \right) dx &= \frac{x}{T^3} + C
,\\
&\vdots
\end{align}
$$
With the help of this, we can easily compute things like
$$
\begin{align}
\int \left( \frac{1}{T^3} + \frac{5}{T^2} - \frac{2}{T} \right) dx
&= \int \left( \frac14 \left( \frac{4}{T^3} - \frac{3}{T^2} \right) + \frac{\frac34 + 5}{T^2} - \frac{2}{T} \right) dx
\\
&= \int \left( \frac14 \left( \frac{4}{T^3} - \frac{3}{T^2} \right) + \frac{23}{4} \cdot \frac12 \left( \frac{2}{T^2} - \frac{1}{T^1} \right) + \frac{\frac{23}{8}-2}{T} \right) dx
\\
&= \frac14 \frac{x}{T^2} + \frac{23}{8} \frac{x}{T} + \frac{7}{8} \arctan x + C
.
\end{align}
$$
Of course, the relation that we are using, $2n \, I_{n+1} - (2n-1) \, I_n = \frac{x}{T^n}$, really is the reduction formula in disguise. However, the trick is:
(a) to derive the formula just by differentiation (instead of starting with an integral where the exponent is one step lower than the one that we're interested in, inserting a factor of $1$, integrating by parts, and so on),
and
(b) to leave the formula in its "natural" form as it appears when differentiating (instead of solving for $I_{n+1}$ in terms of $I_n$), which results in a structure which is easier to remember and a more pleasant way of organizing the computations.
