A) \[\frac{1}{2}({{x}^{2}}+1){{e}^{{{x}^{2}}}}+c\]
B) \[({{x}^{2}}+1){{e}^{{{x}^{2}}}}+c\]
C) \[\frac{1}{2}({{x}^{2}}-1){{e}^{{{x}^{2}}}}+c\]
D) \[({{x}^{2}}-1){{e}^{{{x}^{2}}}}+c\]
Correct Answer: C
Solution :
Put \[{{x}^{2}}=t\Rightarrow 2x\,dx=dt,\] then \[\int_{{}}^{{}}{{{x}^{3}}{{e}^{{{x}^{2}}}}dx}=\frac{1}{2}\int_{{}}^{{}}{t{{e}^{t}}dt}\] \[=\frac{1}{2}\left[ t{{e}^{t}}-{{e}^{t}} \right]+c\]\[=\frac{1}{2}{{e}^{{{x}^{2}}}}({{x}^{2}}-1)+c.\]
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