A) \[-2{{x}^{3}}-1\]
B) \[-\text{ }2{{x}^{3}}+1\]
C) \[4{{x}^{3}}+1\]
D) \[-\,4{{x}^{3}}-1\]
Correct Answer: D
Solution :
\[\int{{{x}^{5}}.{{e}^{-\,4{{x}^{3}}}}\,dx\,=\,\,\int{{{x}^{3}}.{{x}^{2}}\,{{e}^{-\,4{{x}^{3}}}}\,dx}}\] Let \[4{{x}^{3}}\,\,=\,\,t\] \[12{{x}^{2}}\,dx=dt\] \[=\,\,\,\int{\frac{1}{4}.\frac{1}{12}{{e}^{-t}}\,dt\,\,=\,\,\frac{1}{48}\int{\underset{I}{\mathop{t}}\,.{{\underset{II}{\mathop{e}}\,}^{-t}}\,dt}}\] \[=\,\,\,\frac{1}{48}\left[ \int{-t.{{e}^{-t}}+\int{{{e}^{-t}}}dt} \right](integration\,\,by\,\,part)\] \[=\,\,\,\frac{1}{48}(-t{{e}^{-t}}-{{e}^{-t}})\] \[=\,\,\,\,-\frac{1}{48}{{e}^{-t}}(t\,\,+\,\,1)\] \[=\,\,\,-\frac{1}{48}{{e}^{-4{{x}^{3}}}}(4{{x}^{3}}+1)\] \[\because \,\,\,f\left( x \right)=-\left( 4{{x}^{3}}+1 \right)=-\text{ }4{{x}^{3}}-1\]
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