Evaluate the following integrals:

Put $3 \log x=t$

We have $\mathrm{d}(\log \mathrm{x})=1 / \mathrm{x}$

Hence, $\mathrm{d}(3 \log \mathrm{x})=\mathrm{dt}=3 / \mathrm{x} \mathrm{dx}$

Or $1 / x d x=d t / 3$

Hence, $\int \frac{1}{x \sqrt{4-9(\log x)^{2}}} d x=\int \frac{1}{3} \frac{d t}{\sqrt{2^{2}-t^{2}}}$

Since we have, $\int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+c$

Hence, $\int \frac{1}{3} \frac{\mathrm{dt}}{\sqrt{2^{2}-\mathrm{t}^{2}}}=\frac{1}{3} \sin ^{-1}\left(\frac{\mathrm{t}}{2}\right)+\mathrm{c}$

Put $t=3 \log x$

$=\frac{1}{3} \sin ^{-1}\left(\frac{t}{2}\right)+c=\frac{1}{3} \sin ^{-1}\left(\frac{3 \log x}{2}\right)+c$