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: 
 A: Notice, your mistake $\sqrt{\text{(Hyp)}^2}=\sqrt{16+x^2}\ne 4+x$
you should have $$\tan\theta=\frac{x}{4}\ \  \text{&} \ \ \sec\theta=\frac{\text{Hyp}}{\text{base}}=\frac{\sqrt{16+x^2}}{4}$$
hence substituting the values of $\sec\theta$ & $\tan\theta$,  one should get 
$$\ln|\sec\theta+\tan\theta|+C=\ln\left|\frac{\sqrt{16+x^2}}{4}+\frac{x}{4}\right|+C$$
$$=\ln\left|x+\sqrt{16+x^2}\right|-\ln (4)+C=\ln\left|x+\sqrt{16+x^2}\right|+c$$
A: I prefer to use this substitution
$$\begin{align}
x &= 4 \sinh u \\
dx &= 4 \cosh u \, du
\end{align}$$
and hence the integral becomes
$$\begin{align}
I &= \int \frac{1}{\sqrt{16(1+\sinh^2u)}} 4 \cosh u du\\
&= \int \frac{4 \cosh u}{4\sqrt{\cosh^2u}}  du \\
&= \int 1 \, du \\
&= u+C \\
&= \sinh^{-1} \frac{x}{4} + C\\
&= \ln \left( \sqrt{(1+(\frac{x}{4})^2}+\frac{x}{4} \right) + C
\end{align}$$
A: HINT: set $x=4\sinh(t)$ and you will get $$16\sinh(t)^2+16=16(\sinh(t)+1)=16\cosh(t)^2$$
A: $u = \ln \left| {x + \sqrt {{x^2} + 16} } \right|$. Then $du = \frac{1}
{{\sqrt {{x^2} + 16} }}dx$. Then
$$\int {\frac{1}{{\sqrt {{x^2} + 16} }}dx}  = \int {du}  = u + C = \ln \left| {x + \sqrt {{x^2} + 16} } \right| + C.$$
