A) \[-2{{x}^{3}}-1\]
B) \[-4{{x}^{3}}-1\]
C) \[-2{{x}^{3}}+1\]
D) \[4{{x}^{3}}+1\]
Correct Answer: B
Solution :
\[\int{{{x}^{5}}.{{e}^{-4{{x}^{3}}}}dx}\] |
\[=\int{{{x}^{2}}.{{x}^{3}}{{e}^{-4{{x}^{3}}}}dx}\] |
\[-4{{x}^{3}}=t\] |
\[-12{{x}^{2}}dx=dt\] |
\[=\frac{-1}{12}\int{-\frac{t}{4}{{e}^{t}}dt}\] |
\[=\frac{1}{48}\int{t{{e}^{t}}dt}\] |
\[=\frac{1}{48}t{{e}^{t}}-1.{{e}^{t}}+c\] |
\[=\frac{1}{48}{{e}^{-4{{x}^{3}}}}..(-4{{x}^{3}})-{{e}^{-4{{x}^{3}}}}+c.\] |
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