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Question:

If $\int x^{5} \mathrm{e}^{-4 x^{3}} d x=\frac{1}{48} \mathrm{e}^{-4 x^{3}} f(x)+\mathrm{C}$, where $\mathrm{C}$ is a constant of integration, then $f(x)$ is equal to:

  1. (1) $-2 x^{3}-1$

  2. (2) $-4 x^{3}-1$

  3. (3) $-2 x^{3}+1$

  4. (4) $4 x^{3}+1$


Correct Option: , 2

Solution:

$I=\int x^{5} e^{-4 x^{3}} d x$

Put $-4 x^{3}=\theta$

$\Rightarrow \quad-12 x^{2} d x=d \theta$

$\Rightarrow x^{2} d x=-\frac{d \theta}{12}$

$I=\int \frac{1}{48} \theta e^{\theta} d \theta=\frac{1}{48}\left[\theta e^{\theta}-e^{\theta}\right]+C$

$I=\frac{1}{48} e^{-4 x^{3}}\left(-4 x^{3}-1\right)+C$

Then, by comparison'

$f(x)=-4 x^{3}-1$

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