A) \[{{I}^{2}}\]
B) \[\frac{1}{2}I\]
C) \[2I\]
D) \[\frac{1}{2}{{I}^{2}}\]
Correct Answer: B
Solution :
Let\[{{I}_{1}}=\int_{0}^{1}{\frac{x{{e}^{{{x}^{2}}}}}{{{x}^{2}}+1}dx}\], putting\[{{x}^{2}}=t\] \[\Rightarrow \] \[2x\,\,dx=dt\], we get \[\therefore \] \[{{I}_{1}}=\frac{1}{2}\int_{0}^{1}{\frac{{{e}^{t}}dt}{t+1}}=\frac{1}{2}\int_{0}^{1}{\frac{{{e}^{x}}dx}{x+1}=\frac{1}{2}I}\] \[\left[ \because \,\,I=\int_{0}^{1}{\frac{{{e}^{x}}dx}{x+1}(given)} \right]\]You need to login to perform this action.
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