Question

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

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

Answer 1

The correct option is A $-4x^3-1$

$\int x^5\cdot e^{-4x^3}dx=\frac{1}{48}{e}^{-4x^3}f\left(x\right)+c$

Put $x^3=t$

$3x^{2}dx=dt$

$\int x^3\cdot e^{-4x^3}\cdot x^{2}dx$

$\frac{1}{3}\int t\cdot e^{-4t}dt$

$\frac{1}{3}[t\cdot \frac{e^{-4t}}{-4}-\int \frac{e^{-4t}}{-4}dt]$

$-\frac{e^{4t}}{48}[4t+1]+c$

$\frac{-e^{-4x^3}}{48}[4x^3+1]+c$

$\therefore f\left(x\right)=-1-4x^3$

From the given options (1) is most suitable