Question

Evaluate the following integrals:
$\int \frac{\sqrt{16+{\left(\log x\right)}^2}}{x}dx$

Answer 1

Let $I=\int \frac{\sqrt{16+(\log x{)}^2}}{x}dx$

$⟹I=\int \sqrt{16+(\log x{)}^2}\times \frac{dx}{x}$

Put $\log x=t⟹\frac{dx}{x}=dt$

$⟹I=\int \sqrt{4^2+t^2}dt$

As we know that

$\int \sqrt{a^2+x^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\ln |x+\sqrt{x^2+a^2}|+C$

Here $a=4$

$⟹I=\frac{t}{2}\sqrt{t^2+4^2}+\frac{16}{2}\ln |t+\sqrt{t^2+4^2}|+C$

$⟹I=\frac{\log x}{2}\sqrt{(\log x{)}^2+4^2}+8\ln |\log x+\sqrt{(\log x{)}^2+4^2}|+C$ $(\because t=\log x)$

$⟹I=\frac{\log x}{2}\sqrt{(\log x{)}^2+16}+8\ln |\log x+\sqrt{(\log x{)}^2+16}|+C$