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:
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'
$f(x)=-4 x^{3}-1$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.