Question

$\int \frac{\log \sqrt{x}}{3x}dx$ is equal to

• A. $\frac{1}{3}(\log \sqrt{x}{)}^2+c$
• B. $\frac{2}{3}(\log \sqrt{x}{)}^2+c$
• C. $\frac{2}{3}(\log x{)}^2+c$
• D. $\frac{1}{3}(\log x{)}^2+c$

Answer 1

The correct option is A $\frac{1}{3}(\log \sqrt{x}{)}^2+c$

$\int \frac{\log \sqrt{x}}{3x}dx$
$\log \sqrt{x}=t\frac{1}{2x}dx=dt\Rightarrow \int \frac{2t}{3}dt=\frac{t^2}{3}=\frac{1}{3}(\log \sqrt{x}{)}^2+c$