Question

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

• A. $ - 4 x^3 - 1 $
• B. $ 4 x^3 + 1 $
• C. $ - 2 x^3 - 1 $
• D. $ - 2 x^3 + 1 $

Answer 1

Answer: (A)

$\int x^{5} .e^{-4x^3} dx = \frac{1}{48} e^{-4x^3}f\left(x\right)+c $
Put $ x^{3}=t $
$3x^{2} dx =dt$
$ \int x^{3}.e^{-4x^3}.x^{2}dx$
$ \frac{1}{3} \int t.e^{-4t}dt $
$ \frac{1}{3} \left[t. \frac{e^{-4t}}{-4} -\int \frac{e^{-4t}}{-4} dt\right]$
$ - \frac{e^{-4t}}{48} \left[4t+1\right]+c $
$ \frac{-e^{-4x^3}}{48} \left[4x^{3} +1\right]+c $
$\therefore \; f(x) = -1 - 4x^3$
(From the given options (1) is most suitable)