WBJEE · Maths · Indefinite Integration
\(\int \frac{\log \sqrt{x}}{3 x} d x\) 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 & Solution
Correct Answer
(A) \(\frac{1}{3}(\log \sqrt{x})^{2}+C\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \frac{\log \sqrt{x}}{3 x} d x\) Aqain, let \(\log \sqrt{x}=z \Rightarrow \frac{1}{2 x} d x=d z\) \(I=\int \frac{2 z}{3} d z=\frac{2}{3} \int z d z\) \(=\frac{2}{3} \cdot \frac{z^{2}}{2}+C=\frac{1}{3} \left(\log \sqrt{x}\right)^{2}+C\)
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