CUET · MATHS · PYQ PAPER 2025
The value of the integral \(I=\int_0^1 \frac{1}{\sqrt{x^2+2 x+3}} d x\) is :
- A \(\log \left(\frac{2+\sqrt{6}}{1-\sqrt{3}}\right)\)
- B \(\log \left(\frac{2+\sqrt{6}}{1+\sqrt{3}}\right)\)
- C \(\log \left(\frac{5+\sqrt{6}}{1+\sqrt{3}}\right)\)
- D \(\log (5+\sqrt{6})\)
Answer & Solution
Correct Answer
(B) \(\log \left(\frac{2+\sqrt{6}}{1+\sqrt{3}}\right)\)
Step-by-step Solution
Detailed explanation
\(I=\int_0^1 \frac{1}{\sqrt{(x+1)^2+(\sqrt{2})^2}} d x\) \(I=\left[\log \left|(x+1)+\sqrt{(x+1)^2+2}\right|\right]_0^1\) \(I=\left[\log \left|(x+1)+\sqrt{x^2+2x+3}\right|\right]_0^1\) \(I=\log \left|(1+1)+\sqrt{1^2+2(1)+3}\right| - \log \left|(0+1)+\sqrt{0^2+2(0)+3}\right|\)…
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