Choose $u$ using LIATE: Logarithm, Inverse trig, Algebraic, Trig, Exponential.
$\sqrt{a^2-x^2}$: let $x=a\sin\theta$.
$\sqrt{a^2+x^2}$: let $x=a\tan\theta$.
$\sqrt{x^2-a^2}$: let $x=a\sec\theta$.
Decompose rational functions. For repeated factor $(x-a)^n$, include terms $\frac{A_1}{x-a}+\cdots+\frac{A_n}{(x-a)^n}$.
$\int \ln x\,dx = x\ln x - x + C$
$\int \sec^3 x\,dx = \tfrac{1}{2}\sec x\tan x + \tfrac{1}{2}\ln|\sec x+\tan x|+C$
Evaluate $\displaystyle\int x e^x\,dx$.
$\int_a^\infty f(x)\,dx = \lim_{t\to\infty}\int_a^t f(x)\,dx$. Converges if limit is finite.
Converges if $p>1$; Diverges if $p\le 1$.
Does $\displaystyle\int_1^\infty \frac{1}{x^2}\,dx$ converge?
$L=\displaystyle\int_a^b\sqrt{1+[f'(x)]^2}\,dx$
Disk: $V=\pi\displaystyle\int_a^b [f(x)]^2 dx$
Washer: $\pi\displaystyle\int_a^b\!\left([R(x)]^2-[r(x)]^2\right)dx$
$V=2\pi\displaystyle\int_a^b x\,f(x)\,dx$ (rotation about $y$-axis)
$S=2\pi\displaystyle\int_a^b f(x)\sqrt{1+[f'(x)]^2}\,dx$
$\displaystyle\sum_{n=0}^\infty ar^n = \frac{a}{1-r}$ if $|r|<1$; diverges if $|r|\ge 1$.
$\displaystyle\sum_{n=1}^\infty \frac{1}{n^p}$ converges iff $p>1$.
$L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$: converges if $L<1$; diverges if $L>1$; inconclusive if $L=1$.
$\sum(-1)^n b_n$ converges if (1) $b_n$ decreasing, (2) $b_n\to 0$.
For $\sum a_n(x-c)^n$: $R=\dfrac{1}{\limsup|a_n|^{1/n}}$. Use Ratio Test to find $R$.
$e^x=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}$, $\sin x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^n x^{2n+1}}{(2n+1)!}$,
$\cos x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^n x^{2n}}{(2n)!}$
Find the Maclaurin series for $e^{2x}$.
$L=\displaystyle\int_\alpha^\beta\sqrt{\!\left(\frac{dx}{dt}\right)^2\!+\!\left(\frac{dy}{dt}\right)^2}\,dt$
$A=\dfrac{1}{2}\displaystyle\int_\alpha^\beta [r(\theta)]^2\,d\theta$
$x=r\cos\theta$, $y=r\sin\theta$, $r^2=x^2+y^2$
$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$
Rewrite as $g(y)\,dy = f(x)\,dx$, then integrate both sides.
$y'+P(x)y=Q(x)$. Integrating factor: $\mu=e^{\int P\,dx}$. Solution: $y=\dfrac{1}{\mu}\displaystyle\int\mu Q\,dx$.
Solve $\dfrac{dy}{dx}=xy$.