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Q.

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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a

$2 I$

b

$\frac{1}{2} I$

c

$I^{2}$

d

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

answer is B.

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Detailed Solution

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

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

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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