A) \[-\frac{{{e}^{{{x}^{2}}}}}{2}+c\]
B) \[\frac{{{e}^{{{x}^{2}}}}}{2}+c\]
C) \[\frac{{{e}^{x}}}{2}+c\]
D) \[-\frac{{{e}^{x}}}{2}+c\]
Correct Answer: B
Solution :
\[I=\int{x{{e}^{{{x}^{2}}}}dx}\]\[=\frac{1}{2}\int{{{e}^{t}}dt=\frac{{{e}^{{{x}^{2}}}}}{2}}+c\]. {Put \[{{x}^{2}}=t\] Þ \[2xdx=dt\]}You need to login to perform this action.
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