Question:
If $\int x^{5} e^{-4 x^{3}} d x=\frac{1}{48} e^{-4 x^{3}} f(x)+C$, where $C$ is a
constant of integration, then $\mathrm{f}(\mathrm{x})$ is equal to:
Correct Option: 1
Solution:
$\int x^{5} \cdot e^{-4 x^{3}} d x=\frac{1}{48} e^{-4 x^{3}} f(x)+c$
Put $x^{3}=t$
$3 x^{2} d x=d t$
$\int x^{3} \cdot e^{-4 x^{3}} \cdot x^{2} d x$
$\frac{1}{3} \int t \cdot e^{-4 t} d t$
$\frac{1}{3}\left[t \cdot \frac{e^{-4 t}}{-4}-\int \frac{e^{-4 t}}{-4} d t\right]$
$-\frac{e^{-4 t}}{48}[4 t+1]+c$
$\frac{-e^{-4 x^{3}}}{48}\left[4 x^{3}+1\right]+c$
$\therefore f(x)=-1-4 x^{3}$
Option (1)
(From the given options (1) is most suitable)
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