MHT CET · Maths · Indefinite Integration
\(\int \frac{\log \sqrt{x}}{3 x} \mathrm{~d} x\) is equal to
- A \(\frac{1}{3}(\log \sqrt{x})+\mathrm{c}\), (where c is a constant of integration)
- B \(\frac{2}{3}(\log \sqrt{x})^2+\mathrm{c}\), (where c is a constant of integration)
- C \(\frac{2}{3}(\log x)^2+\mathrm{c}\), (where c is a constant of integration)
- D \(\frac{1}{12}(\log x)^2+\mathrm{c},(\) where c is a constant of integration)
Answer & Solution
Correct Answer
(D) \(\frac{1}{12}(\log x)^2+\mathrm{c},(\) where c is a constant of integration)
Step-by-step Solution
Detailed explanation
\(\text {Put } x=\mathrm{t}^2 \Rightarrow \mathrm{~d} x=2 \mathrm{tdt} \)
\( \int \frac{\log \sqrt{x}}{3 x} \mathrm{~d} x =\int \frac{\log \mathrm{t}}{3 \mathrm{t}^2}(2 \mathrm{tdt}) \)
\( =\frac{2}{3} \int \frac{\log \mathrm{t}}{\mathrm{t}} \mathrm{dt} \)
\( =\frac{2}{3} \cdot \frac{(\log \mathrm{t})^2}{2}+\mathrm{c}=\frac{(\log \sqrt{x})^2}{3}+\mathrm{c} \)
\( =\frac{1}{3}\left(\frac{1}{2} \log x\right)^2+\mathrm{c} \)
\( =\frac{1}{12}(\log x)^2+\mathrm{c}\)
\( \int \frac{\log \sqrt{x}}{3 x} \mathrm{~d} x =\int \frac{\log \mathrm{t}}{3 \mathrm{t}^2}(2 \mathrm{tdt}) \)
\( =\frac{2}{3} \int \frac{\log \mathrm{t}}{\mathrm{t}} \mathrm{dt} \)
\( =\frac{2}{3} \cdot \frac{(\log \mathrm{t})^2}{2}+\mathrm{c}=\frac{(\log \sqrt{x})^2}{3}+\mathrm{c} \)
\( =\frac{1}{3}\left(\frac{1}{2} \log x\right)^2+\mathrm{c} \)
\( =\frac{1}{12}(\log x)^2+\mathrm{c}\)
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