Mnemonics for typical integrations I am supposed to mug up these integrals for my upcoming exams:
$$\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac xa+C$$
$$\int\sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\ln|x+\sqrt{x^2+a^2}|+C$$
$$\int\sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\ln|x+\sqrt{x^2-a^2}|+C$$
I have been trying all my past one year trying to mug them up, ever since this chapter of indefinite integration was introduced, yet I keep messing the terms/$\pm$ signs/etc. every single time.
I hope there could be a few mnemonics to help learn these integrals. I am not necessarily looking for a song or an animated gif, but perhaps, any thing that the mind can instantly hook up to, and connect to these equations, is appreciated. 

As bames suggested above, the method of solving for these indefinite integrations is actually a helpful tool in quickly recalling them. Here, I will provide a simple derivation - for the three integrations I listed - that helped me recall them within thirty seconds on paper. This is a long post, because it captures the exact method that I currently use.

$\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac xa+C$
$$\begin{align}
I=\int\sqrt{a^2-x^2}~dx
&=\int\frac{a^2-x^2}{\sqrt{a^2-x^2}}dx\\
&=\int\frac{a^2}{\sqrt{a^2-x^2}}dx-\int\frac{x^2}{\sqrt{a^2-x^2}}dx\\
&=a^2\sin^{-1}\frac xa-\int x\cdot\frac{x}{\sqrt{a^2-x^2}}dx
\end{align}$$
Now, apply the Integration-By-Parts rule on the second integrand as follows:
$$
\begin{align}
\int x\cdot\frac{x}{\sqrt{a^2-x^2}}dx
&=x\int\frac{x}{\sqrt{a^2-x^2}}-\int\int\frac{x}{\sqrt{a^2-x^2}}dx\\
&=-x\sqrt{a^2-x^2}+\int \sqrt{a^2-x^2}dx\\
&=-x\sqrt{a^2-x^2}+I
\end{align}$$
Substituting it back, we get:
$$\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac xa+C$$

$\int\sqrt{x^2+a^2}~dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\ln|x+\sqrt{x^2+a^2}|+C$
$$
\begin{align}
I=\int\sqrt{x^2+a^2}~dx
&=\int\frac{x^2+a^2}{\sqrt{x^2+a^2}}dx\\
&=\int\frac{x^2}{\sqrt{x^2+a^2}}dx+\int\frac{a^2}{\sqrt{x^2+a^2}}dx\\
&=\int x\cdot\frac{x}{\sqrt{x^2+a^2}}dx+a^2\ln|\sqrt{x^2+a^2}+x|+C'
\end{align}
$$
Now, apply the Integration-By-Parts rule on the first integrand as follows:
$$
\begin{align}
\int x\cdot\frac{x}{\sqrt{x^2+a^2}}dx&=x\int\frac{x}{\sqrt{x^2+a^2}}dx-\int\int \frac x{\sqrt{x^2+a^2}}dx\\
&=x\sqrt{x^2+a^2}-\int\sqrt{x^2+a^2}\\
&=x\sqrt{x^2+a^2}-I
\end{align}$$
Substituting it back, we get: $$I=\frac x2\sqrt{x^2+a^2}+\frac{a^2}2\ln|\sqrt{x^2+a^2}+x|+C$$

$\int\sqrt{x^2-a^2}~dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\ln|x+\sqrt{x^2-a^2}|+C$
$$
\begin{align}
I=\int\sqrt{x^2-a^2}~dx
&=\int\frac{x^2-a^2}{\sqrt{x^2-a^2}}dx\\
&=\int\frac{x^2}{\sqrt{x^2-a^2}}dx-\int\frac{a^2}{\sqrt{x^2-a^2}}dx\\
&=\int x\cdot\frac{x}{\sqrt{x^2-a^2}}dx-a^2\ln|\sqrt{x^2-a^2}+x|+C'
\end{align}
$$
Now, apply the Integration-By-Parts rule on the first integrand as follows:
$$
\begin{align}
\int x\cdot\frac{x}{\sqrt{x^2-a^2}}dx
&=x\int\frac{x}{\sqrt{x^2-a^2}}dx-\int\int \frac x{\sqrt{x^2-a^2}}dx\\
&=x\sqrt{x^2-a^2}-\int\sqrt{x^2-a^2}\\
&=x\sqrt{x^2-a^2}-I
\end{align}$$
Substituting it back, we get: $$I=\frac x2\sqrt{x^2-a^2}-\frac{a^2}2\ln|\sqrt{x^2-a^2}+x|+C$$

Quick references:
$$\begin{align}
\int\frac1{\sqrt{x^2+a^2}}dx&=\frac 1{a}\int\frac{a\sec^2\theta}{\sqrt{\tan^2\theta+1}}d\theta ~~(\because x=a\tan\theta)\\
&=\int\sec\theta d\theta\\
&=\ln|\sec\theta+\tan\theta|+C'\\
&=\ln|\sqrt{x^2+a^2}+x|+C
\end{align}$$

$$\begin{align}
\int\frac1{\sqrt{x^2-a^2}}dx&=\frac 1{a}\int\frac{a\sec\theta\tan\theta}{\sqrt{\sec^2\theta-1}}d\theta ~~(\because x=a\sec\theta)\\
&=\int\sec\theta d\theta\\
&=\ln|\sec\theta+\tan\theta|+C'\\
&=\ln|\sqrt{x^2-a^2}+x|+C
\end{align}$$

$$\begin{align}
\int\frac1{\sqrt{a^2-x^2}}dx&=
\frac 1{a}\int\frac{a\cos\theta}{\sqrt{1-\sin^2\theta}}d\theta ~~(\because x=a\sin\theta)\\
&=\int d\theta\\
&=\theta+C'\\
&=\sin^{-1}\frac xa+C
\end{align}$$
