$\int \frac{\log \sqrt{x}}{3 x} \mathrm{~d} x$ is equal to

$\int \frac{\log \sqrt{x}}{3 x} \mathrm{~d} x$ is equal to
  1. $\frac{1}{3}(\log \sqrt{x})+\mathrm{c}$, (where c is a constant of integration)
  2. $\frac{2}{3}(\log \sqrt{x})^2+\mathrm{c}$, (where c is a constant of integration)
  3. $\frac{2}{3}(\log x)^2+\mathrm{c}$, (where c is a constant of integration)
  4. $\frac{1}{12}(\log x)^2+\mathrm{c},($ where c is a constant of integration)

Solution

$\begin{aligned} & \text { Put } x=\mathrm{t}^2 \Rightarrow \mathrm{~d} x=2 \mathrm{tdt} \\ & \qquad \begin{aligned} \int \frac{\log \sqrt{x}}{3 x} \mathrm{~d} x & =\int \frac{\log \mathrm{t}}{3 \mathrm{t}^2}(2 \mathrm{tdt}) \\ & =\frac{2}{3} \int \frac{\log \mathrm{t}}{\mathrm{t}} \mathrm{dt} \\ & =\frac{2}{3} \cdot \frac{(\log \mathrm{t})^2}{2}+\mathrm{c}=\frac{(\log \sqrt{x})^2}{3}+\mathrm{c} \\ & =\frac{1}{3}\left(\frac{1}{2} \log x\right)^2+\mathrm{c} \\ & =\frac{1}{12}(\log x)^2+\mathrm{c}\end{aligned}\end{aligned}$

Asked in: MHT CET 2024 (02 May Shift 2)

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