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. $\frac{2}{3}\left(log\,x\right)^{2}+c$
• D. $\frac{1}{3}\left(log\,x\right)^{2}+c$

Answer 1

Answer: (A)

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