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JEE Main 20249 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2024 · 9 April · Evening Shift · Mathematics · Definite Integration

The value of the integral ∫_ -1 ^2 _e(x+ x^2+1 ) d x is

The value of the integral ∫−12log⁡e(x+x2+1)dx\int_{-1}^2 \log _e\left(x+\sqrt{x^2+1}\right) d x is

Options

  1. A5−2+log⁡e(7+451+2)\sqrt{5}-\sqrt{2}+\log _e\left(\frac{7+4 \sqrt{5}}{1+\sqrt{2}}\right)
  2. B5−2+log⁡e(9+451+2)\sqrt{5}-\sqrt{2}+\log _e\left(\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right)
  3. C2−5+log⁡e(7+451+2)\sqrt{2}-\sqrt{5}+\log _e\left(\frac{7+4 \sqrt{5}}{1+\sqrt{2}}\right)
  4. D2−5+log⁡e(9+451+2)\sqrt{2}-\sqrt{5}+\log _e\left(\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right)

Correct answer

D. 2−5+log⁡e(9+451+2)\sqrt{2}-\sqrt{5}+\log _e\left(\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right)

Step-by-step solution

I=∫−12log⁡e(x+x2+1)dx=xlog⁡e(x+x2+1)−∫−12(1+xx2+1x+x2+1)dx=xlog⁡e(x+x2+1)−∫−12xx2+1dx\begin{aligned} & I=\int_{-1}^2 \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\end{aligned}=xlog⁡e(x+x2+1)−x2+1∣−12=(2log⁡e(2+5)−5)−(−log⁡e(−1+2)−2)=log⁡e(2+5)2−5+log⁡e(2−1)+2=log⁡e(2+5)2−5+log⁡e(2−1)+2\begin{aligned} & =x \log _e\left(x+\sqrt{x^2+1}\right)-\left.\sqrt{x^2+1}\right|_{-1} ^2 \\ & =\left(2 \log _e(2+\sqrt{5})-\sqrt{5}\right) \\ & -\left(-\log _e(-1+\sqrt{2})-\sqrt{2}\right) \\ & =\log _e(2+\sqrt{5})^2-\sqrt{5}+\log _e(\sqrt{2}-1)+\sqrt{2} \\ & =\log _e(2+\sqrt{5})^2-\sqrt{5}+\log _e(\sqrt{2}-1)+\sqrt{2}\end{aligned} =2−5+log⁡e((2+5)22+1)=2−5+log⁡e(9+452+1)\begin{aligned} & =\sqrt{2}-\sqrt{5}+\log _e\left(\frac{(2+\sqrt{5})^2}{\sqrt{2+1}}\right) \\ & =\sqrt{2}-\sqrt{5}+\log _e\left(\frac{9+4 \sqrt{5}}{\sqrt{2+1}}\right)\end{aligned}$
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