<s>[INST] Evaluate the Integral: $\int\frac{dx}{\sqrt{x^2+16}}$ I want to evaluate $\int\frac{dx}{\sqrt{x^2+16}}$.
My answer is: $\ln \left| \frac{4+x}{4}+\frac{x}{4} \right|+C$
My work is attached: 
 [/INST] I prefer to use this substitution
$$\begin{align}
x &= 4 \sinh u \\
dx &= 4 \cosh u \, du
\end{align}$$
and hence the integral becomes
$$\begin{align}
I &= \int \frac{1}{\sqrt{16(1+\sinh^2u)}} 4 \cosh u du\\
&= \int \frac{4 \cosh u}{4\sqrt{\cosh^2u}}  du \\
&= \int 1 \, du \\
&= u+C \\
&= \sinh^{-1} \frac{x}{4} + C\\
&= \ln \left( \sqrt{(1+(\frac{x}{4})^2}+\frac{x}{4} \right) + C
\end{align}$$
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