• # question_answer If$\int_{{}}^{{}}{{{x}^{5}}{{e}^{-4{{x}^{3}}}}dx=\frac{1}{48}{{e}^{-4{{x}^{3}}}}}f(x)+C,$ where C is a constant of integration, then$f(x)$ is equal to: A) $-2{{x}^{3}}-1$ B) $-4{{x}^{3}}-1$ C) $-2{{x}^{3}}+1$ D) $4{{x}^{3}}+1$

 $\int{{{x}^{5}}.{{e}^{-4{{x}^{3}}}}dx}$ $=\int{{{x}^{2}}.{{x}^{3}}{{e}^{-4{{x}^{3}}}}dx}$ $-4{{x}^{3}}=t$ $-12{{x}^{2}}dx=dt$ $=\frac{-1}{12}\int{-\frac{t}{4}{{e}^{t}}dt}$ $=\frac{1}{48}\int{t{{e}^{t}}dt}$ $=\frac{1}{48}t{{e}^{t}}-1.{{e}^{t}}+c$ $=\frac{1}{48}{{e}^{-4{{x}^{3}}}}..(-4{{x}^{3}})-{{e}^{-4{{x}^{3}}}}+c.$