JEE Mains · Maths · STD 12 - 7.1 indefinite integral
The value of the integral \(\int_{-1}^2 \log _e\left(x+\sqrt{x^2+1}\right) d x\) is :
- A \(\sqrt{5}-\sqrt{2}+\log _{\mathrm{e}}\left(\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right)\)
- B \(\sqrt{2}-\sqrt{5}+\log _{\mathrm{e}}\left(\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right)\)
- C \(\sqrt{5}-\sqrt{2}+\log _{\mathrm{e}}\left(\frac{7+4 \sqrt{5}}{1+\sqrt{2}}\right)\)
- D \(\sqrt{2}-\sqrt{5}+\log _{\mathrm{e}}\left(\frac{7+4 \sqrt{5}}{1+\sqrt{2}}\right)\)
Answer & Solution
Correct Answer
(B) \(\sqrt{2}-\sqrt{5}+\log _{\mathrm{e}}\left(\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right)\)
Step-by-step Solution
Detailed explanation
\( I=\int_{-1}^2 1 \log _e\left(x+\sqrt{x^2+1}\right) d x \) \( =x \log _e\left(x+\sqrt{x^2+1}\right)-\int_{-1}^2\left(\frac{1+\frac{x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}}\right) d x \) \( =x \log _e\left(x+\sqrt{x^2+1}\right)-\int_{-1}^2 \frac{x}{\sqrt{x^2+1}} d x \)…
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