Question

\(\int \frac{\log \sqrt{x}}{3 x} d x\) 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

Ans: (A)
 \(\int \frac{\log \sqrt{x}}{3 x} d x=I\)
Let \(\log \sqrt{x}=z \Rightarrow \frac{1}{2 x} d x=d z\)
$$
\begin{aligned}
\therefore I=\int \frac{2 z}{3} d z &=\frac{2}{3} \int z d z \\
&=\frac{2}{3} \cdot \frac{z^{2}}{z}+c \\
&=\frac{1}{3} \cdot(\log \sqrt{x})^{2}+c
\end{aligned}
$$