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if $\int x^5e^{-4x^3}dx = \frac{1}{48}e^{-4x^3}f(x) + C$ ,

where C is a constant of integration, then $f(x)$ is equal to
Option 1)
$-2x^3-1$
Option 2)
$-4x^3-1$
Option 3)
$-2x^3+1$
Option 4)
$4x^3+1$

Answers (1)

best_answer

Integration By PARTS -

Let $u$ and $v$ are two functions then

$\int u\cdot vdx=u\int vdx-\int \left ( \frac{du}{dx}\int vdx \right )dx$

- wherein

Where $u$ is the Ist function $v$ is he IInd function

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

put $x^{3}=t$

$3 x^{2}dx=dt$

Now,

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

$=\frac{1}{3}\int te^{-4t}dt$

from the concept of Integration by parts

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

$=-\frac{e^{-4t}}{48}\left [ 4t+1 \right ]+C$

Substituiting back $x^{3}=t$

$\therefore$ $-\frac{e^{-4x^{3}}}{48}\left [ 4x^{3}+1 \right ]+C$

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

Option 1)

$-2x^3-1$

Option 2)
$-4x^3-1$
Option 3)

$-2x^3+1$

Option 4)

$4x^3+1$

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