CUET · MATHS · PYQ PAPER 2025
\(\int \sqrt{1+\frac{x^2}{9}} d x\) is equal to
(Where \(C\) is an arbitrary constant)
- A \(\frac{1}{9}\left[\frac{9}{2} \log \left|x+\sqrt{x^2+9}\right|+\frac{x}{2} \sqrt{x^2+9}\right]+C\)
- B \(3\left[\frac{9}{2} \log \left|x+\sqrt{x^2+9}\right|+\frac{x}{2} \sqrt{x^2+9}\right]+C\)
- C \(\left[\frac{9}{2} \log \left|x+\sqrt{x^2+9}\right|+\frac{x}{2} \sqrt{x^2+9}\right]+C\)
- D \(\frac{1}{3}\left[\frac{9}{2} \log \left|x+\sqrt{x^2+9}\right|+\frac{x}{2} \sqrt{x^2+9}\right]+C\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{3}\left[\frac{9}{2} \log \left|x+\sqrt{x^2+9}\right|+\frac{x}{2} \sqrt{x^2+9}\right]+C\)
Step-by-step Solution
Detailed explanation
\(\int \sqrt{1+\frac{x^2}{9}} d x = \frac{1}{3} \int \sqrt{x^2+3^2} d x\) \( = \frac{1}{3} \left[ \frac{x}{2}\sqrt{x^2+3^2} + \frac{3^2}{2}\log \left|x+\sqrt{x^2+3^2}\right| \right] + C\)…
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