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

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

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