CUET · MATHS · PYQ PAPER 2025
\(\int \frac{\sqrt{16+(\log x)^2}}{x} d x\) is equal to (where \(C\) is an arbitrary constant)
- A \(2 \log \left|\log x+\sqrt{16+(\log x)^2}\right|+\log x \sqrt{16+(\log x)^2}+C\)
- B \(16 \log \left|\log x+\sqrt{16+(\log x)^2}\right|+\frac{\log x}{2} \sqrt{16+(\log x)^2}+C\)
- C \(8 \log \left|\log x+\sqrt{16+(\log x)^2}\right|+\frac{\log x}{2} \sqrt{16+(\log x)^2}+C\)
- D \(4 \log \left|\log x+\sqrt{16+(\log x)^2}\right|+\frac{\log x}{2} \sqrt{16+(\log x)^2}+C\)
Answer & Solution
Correct Answer
(C) \(8 \log \left|\log x+\sqrt{16+(\log x)^2}\right|+\frac{\log x}{2} \sqrt{16+(\log x)^2}+C\)
Step-by-step Solution
Detailed explanation
Let \(u = \log x\). \(du = \frac{1}{x} dx\). \(\int \sqrt{16+u^2} du = \frac{u}{2}\sqrt{16+u^2} + \frac{16}{2}\log|u+\sqrt{16+u^2}| + C\). \(= \frac{\log x}{2}\sqrt{16+(\log x)^2} + 8\log|\log x+\sqrt{16+(\log x)^2}| + C\).
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