Question

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

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

Answer 1

Answer: (A)

$\int \frac{log\sqrt{x}}{3x}dx=I$
Let log $\sqrt{x}=z\Rightarrow \frac{1}{2x}dx=dz$
$\therefore I=\int \frac{2z}{3}dz=\frac{2}{3}\int zdz$
$=\frac{2}{3}.\frac{z^2}{2}+c$
$=\frac{1}{3}.{\left(log\sqrt{x}\right)}^2+c$