Let I=∫01 exx+1dx then the value of the integral I1=∫01 xex2x2+1dx is

I=∫01 exx+1dxI1=∫01 xex2x2+1dxput x2=t ∴ I1=12∫01 ett+1dt=12I

Let I=∫01 exx+1dx then the value of the integral I1=∫01 xex2x2+1dx is

I=∫01 exx+1dxI1=∫01 xex2x2+1dxput x2=t ∴ I1=12∫01 ett+1dt=12I

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Let $I = \int_{0}^{1} \frac{e^{x}}{x + 1} d x$ then the value of the integral $I_{1} = \int_{0}^{1} \frac{x e^{x^{2}}}{x^{2} + 1} d x$ is

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$2 I$

$\frac{1}{2} I$

$I^{2}$

$\frac{1}{2} I^{2}$

answer is B.

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$I = \int_{0}^{1} \frac{e^{x}}{x + 1} d x$

$I_{1} = \int_{0}^{1} \frac{x e^{x^{2}}}{x^{2} + 1} d x$

put $x^{2} = t$

$∴ I_{1} = \frac{1}{2} \int_{0}^{1} \frac{e^{t}}{t + 1} d t = \frac{1}{2} I$

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Let I=∫01 exx+1dx then the value of the integral I1=∫01 xex2x2+1dx is

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Let $I = \int_{0}^{1} \frac{e^{x}}{x + 1} d x$ then the value of the integral $I_{1} = \int_{0}^{1} \frac{x e^{x^{2}}}{x^{2} + 1} d x$ is