<s>[INST] 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.
 [/INST] 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.
</s>