The value of \int_-1^1 \frac(1+\sqrt|x|-x) e^x+(\sqrt|x|-x) e^-xe^x+e^-x d x is equal to

Definite Integration PYQ — JEE Main 2025 · 4 Apr · Shift 1 · Q34 with Solution

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Correct answer: A

Let $I=\int_{-1}^1 \frac{(1+\sqrt{|x|-x})e^x+(\sqrt{|x|-x})e^{-x}}{e^x+e^{-x}}\,dx.$ We simplify the expression using the value of $|x|$ on the intervals $[-1,0]$ and $[0,1]$ .
For $x\in[0,1]$ , we have $|x|=x$ , so $|x|-x=x-x=0\quad\Rightarrow\quad \sqrt{|x|-x}=0.$ Thus the integrand becomes $\frac{(1+0)e^x+0\cdot e^{-x}}{e^x+e^{-x}}=\frac{e^x}{e^x+e^{-x}}.$ So $I_1=\int_0^1 \frac{e^x}{e^x+e^{-x}}\,dx.$ Multiply numerator and denominator by $e^x$ : $\frac{e^x}{e^x+e^{-x}}=\frac{e^{2x}}{e^{2x}+1}=1-\frac{1}{e^{2x}+1}.$ But a better substitution is $t=e^{2x}$ , yet we will use symmetry later.
For $x\in[-1,0]$ , we have $|x|=-x$ , so $|x|-x=-x-x=-2x,$ therefore $\sqrt{|x|-x}=\sqrt{-2x}.$ Hence the integrand becomes

=\frac{e^x+\sqrt{-2x}(e^x+e^{-x})}{e^x+e^{-x}} =\frac{e^x}{e^x+e^{-x}}+\sqrt{-2x}.$$ So $$I_2=\int_{-1}^0 \frac{e^x}{e^x+e^{-x}}\,dx+\int_{-1}^0 \sqrt{-2x}\,dx.$$ 4. Therefore, $$I=\int_{-1}^1 \frac{e^x}{e^x+e^{-x}}\,dx+\int_{-1}^0 \sqrt{-2x}\,dx.$$ Now simplify the first integral: $$J=\int_{-1}^1 \frac{e^x}{e^x+e^{-x}}\,dx.$$ Multiply numerator and denominator by $e^x$: $$\frac{e^x}{e^x+e^{-x}}=\frac{e^{2x}}{1+e^{2x}}.$$ Let $$f(x)=\frac{e^{2x}}{1+e^{2x}}.$$ Then $$f(-x)=\frac{e^{-2x}}{1+e^{-2x}}=\frac{1}{1+e^{2x}}.$$ So $$f(x)+f(-x)=1.$$ Hence over symmetric limits, $$J=\int_{-1}^1 f(x)\,dx=\int_0^1 [f(x)+f(-x)]\,dx=\int_0^1 1\,dx=1.$$ 5. Now compute $$K=\int_{-1}^0 \sqrt{-2x}\,dx.$$ Let $$u=-2x\quad\Rightarrow\quad dx=-\frac{du}{2}.$$ When $x=-1$, $u=2$; when $x=0$, $u=0$. Thus $$K=\int_2^0 \sqrt{u}\left(-\frac{du}{2}\right)=\frac12\int_0^2 u^{1/2}\,du.$$ So $$K=\frac12\left[\frac{2}{3}u^{3/2}\right]_0^2=\frac{1}{3}(2)^{3/2}=\frac{2\sqrt2}{3}.$$ 6. Therefore, $$I=J+K=1+\frac{2\sqrt2}{3}.$$ 7. Comparing with the options: - A: $1+\frac{2\sqrt2}{3}$ - B: $1-\frac{2\sqrt2}{3}$ - C: $2+\frac{2\sqrt2}{3}$ - D: $3-\frac{2\sqrt2}{3}$ The correct option is **A**.

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Correct answer: A

Let $I=\int_{-1}^1 \frac{(1+\sqrt{|x|-x})e^x+(\sqrt{|x|-x})e^{-x}}{e^x+e^{-x}}\,dx.$ We simplify the expression using the value of $|x|$ on the intervals $[-1,0]$ and $[0,1]$ .
For $x\in[0,1]$ , we have $|x|=x$ , so $|x|-x=x-x=0\quad\Rightarrow\quad \sqrt{|x|-x}=0.$ Thus the integrand becomes $\frac{(1+0)e^x+0\cdot e^{-x}}{e^x+e^{-x}}=\frac{e^x}{e^x+e^{-x}}.$ So $I_1=\int_0^1 \frac{e^x}{e^x+e^{-x}}\,dx.$ Multiply numerator and denominator by $e^x$ : $\frac{e^x}{e^x+e^{-x}}=\frac{e^{2x}}{e^{2x}+1}=1-\frac{1}{e^{2x}+1}.$ But a better substitution is $t=e^{2x}$ , yet we will use symmetry later.
For $x\in[-1,0]$ , we have $|x|=-x$ , so $|x|-x=-x-x=-2x,$ therefore $\sqrt{|x|-x}=\sqrt{-2x}.$ Hence the integrand becomes

=\frac{e^x+\sqrt{-2x}(e^x+e^{-x})}{e^x+e^{-x}} =\frac{e^x}{e^x+e^{-x}}+\sqrt{-2x}.$$ So $$I_2=\int_{-1}^0 \frac{e^x}{e^x+e^{-x}}\,dx+\int_{-1}^0 \sqrt{-2x}\,dx.$$ 4. Therefore, $$I=\int_{-1}^1 \frac{e^x}{e^x+e^{-x}}\,dx+\int_{-1}^0 \sqrt{-2x}\,dx.$$ Now simplify the first integral: $$J=\int_{-1}^1 \frac{e^x}{e^x+e^{-x}}\,dx.$$ Multiply numerator and denominator by $e^x$: $$\frac{e^x}{e^x+e^{-x}}=\frac{e^{2x}}{1+e^{2x}}.$$ Let $$f(x)=\frac{e^{2x}}{1+e^{2x}}.$$ Then $$f(-x)=\frac{e^{-2x}}{1+e^{-2x}}=\frac{1}{1+e^{2x}}.$$ So $$f(x)+f(-x)=1.$$ Hence over symmetric limits, $$J=\int_{-1}^1 f(x)\,dx=\int_0^1 [f(x)+f(-x)]\,dx=\int_0^1 1\,dx=1.$$ 5. Now compute $$K=\int_{-1}^0 \sqrt{-2x}\,dx.$$ Let $$u=-2x\quad\Rightarrow\quad dx=-\frac{du}{2}.$$ When $x=-1$, $u=2$; when $x=0$, $u=0$. Thus $$K=\int_2^0 \sqrt{u}\left(-\frac{du}{2}\right)=\frac12\int_0^2 u^{1/2}\,du.$$ So $$K=\frac12\left[\frac{2}{3}u^{3/2}\right]_0^2=\frac{1}{3}(2)^{3/2}=\frac{2\sqrt2}{3}.$$ 6. Therefore, $$I=J+K=1+\frac{2\sqrt2}{3}.$$ 7. Comparing with the options: - A: $1+\frac{2\sqrt2}{3}$ - B: $1-\frac{2\sqrt2}{3}$ - C: $2+\frac{2\sqrt2}{3}$ - D: $3-\frac{2\sqrt2}{3}$ The correct option is **A**.