Originally Posted by
Soroban Hello, ron007!
If you're not familiar with the arcsecant formula,
. . there's always Trig Substitution.
Let $\displaystyle 3x \,=\,4\sec\theta\quad\Rightarrow\quad x \,=\,\frac{4}{3}\sec\theta\quad\Rightarrow\quad dx\,=\,\frac{4}{3}\sec\theta\tan\theta\,d\theta$
Then: .$\displaystyle \sqrt{9x^2 - 16} \:=\:\sqrt{(3x)^2 - 16} \:=\:\sqrt{(4\sec\theta)^2 - 16} \:=\:\sqrt{16\sec^2\theta-16}$
. . $\displaystyle =\:\sqrt{16(\sec^2\theta - 1)} \:=\:\sqrt{16\tan^2\theta} \:=\:4\tan\theta$
Substitute: .$\displaystyle \int\frac{\frac{4}{3}\sec\theta\tan\theta\,d\theta }{\frac{4}{3}\sec\theta\!\cdot\!4\tan\theta} \;=\;\frac{1}{4}\int d\theta \:=\:\frac{1}{4}\theta + C$
Back-substitute
We have: .$\displaystyle 3x \,=\,4\sec\theta\quad\Rightarrow\quad \sec\theta \,=\,\frac{3x}{4}\quad\Rightarrow\quad \theta\,=\,\text{arcsec}\frac{3x}{4}
$
Therefore, the answer is: .$\displaystyle \frac{1}{4}\text{arcsec}\frac{3x}{4} + C$