The value of ∫_ -1 ^1 (1+ |x|-x ) e^x+( |x|-x ) e^ -x e^x+e^ -x d x is equal to

Correct answer: D. 1+ 2 2 3 aligned & I=∫_ -1 ^1 (1+ |-x|-(-x) ) e^ -x +( |-x|-(-x) ) e^ -(-x) e^ -x +e^ -(-x) dx & I=∫_ -1 ^1 (1+ |x|+x ) e^ -x +( |x|+x ) e^ x e^ x +e^ x dx & 2 I=∫_ -1 ^1 (1+ |x|+x + |x|-x )(e^ x +e^ -x ) (e^ x +e^ -x ) dx & 2 I=∫_ -1 ^1(1+ |x|+x + |x|-x ) dx & 2 I=2 ∫_0^1(1+ |x|+x + |x|-x ) dx & 2 I=2 ∫_0^1(1+ 2 x + 0 ) dx & I=∫_0^1(1+ 2 x ) dx=[x+ 2 2 3 x^ 3 / 2 ]_0^1 & I= 2 2 3 +1 aligned

JEE Main 20254 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 4 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Definite Integration

JEE Main 2025 · 4 April · Morning Shift · Mathematics · Definite Integration

The value of ∫_ -1 ^1 (1+ |x|-x ) e^x+( |x|-x ) e^ -x e^x+e^ -x d x...

The value of

\int_{-1}^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} d x

is equal to

Options

A $3-\frac{2 \sqrt{2}}{3}$
B $2+\frac{2 \sqrt{2}}{3}$
C $1-\frac{2 \sqrt{2}}{3}$
D $1+\frac{2 \sqrt{2}}{3}$

Correct answer

D. $1+\frac{2 \sqrt{2}}{3}$

Step-by-step solution

\begin{aligned} & \mathrm{I}=\int_{-1}^1 \frac{(1+\sqrt{|-\mathrm{x}|-(-\mathrm{x})}) \mathrm{e}^{-\mathrm{x}}+(\sqrt{|-\mathrm{x}|-(-\mathrm{x})}) \mathrm{e}^{-(-\mathrm{x})}}{\mathrm{e}^{-\mathrm{x}}+\mathrm{e}^{-(-\mathrm{x})}} \mathrm{dx} \\ & \Rightarrow \mathrm{I}=\int_{-1}^1 \frac{(1+\sqrt{|\mathrm{x}|+\mathrm{x}}) \mathrm{e}^{-\mathrm{x}}+(\sqrt{|\mathrm{x}|+\mathrm{x}}) \mathrm{e}^{\mathrm{x}}}{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{\mathrm{x}}} \mathrm{dx} \\ & \Rightarrow 2 \mathrm{I}=\int_{-1}^1 \frac{(1+\sqrt{|\mathrm{x}|+\mathrm{x}}+\sqrt{|\mathrm{x}|-\mathrm{x}})\left(\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}\right)}{\left(\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}\right)} \mathrm{dx} \\ & \Rightarrow 2 \mathrm{I}=\int_{-1}^1(1+\sqrt{|\mathrm{x}|+\mathrm{x}}+\sqrt{|\mathrm{x}|-\mathrm{x}}) \mathrm{dx} \\ & \Rightarrow 2 \mathrm{I}=2 \int_0^1(1+\sqrt{|\mathrm{x}|+\mathrm{x}}+\sqrt{|\mathrm{x}|-\mathrm{x}}) \mathrm{dx} \\ & \Rightarrow 2 \mathrm{I}=2 \int_0^1(1+\sqrt{2 \mathrm{x}}+\sqrt{0}) \mathrm{dx} \\ & \Rightarrow \mathrm{I}=\int_0^1(1+\sqrt{2 \mathrm{x}}) \mathrm{dx}=\left[\mathrm{x}+\frac{2 \sqrt{2}}{3} \mathrm{x}^{3 / 2}\right]_0^1 \\ & \Rightarrow \mathrm{I}=\frac{2 \sqrt{2}}{3}+1\end{aligned}

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