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Integration by Parts

If ∫ x 5 e -4...

Question

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

A

- 2 x^{3} - 1

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B

- 4 x^{3} - 1

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C

4 x^{3} + 1

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D

- 2 x^{3} + 1

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Solution

The correct option is B $- 4 x^{3} - 1$
Given,
$\int x^{5} e^{- 4 x^{3}} d x = \frac{1}{48} e^{- 4 x^{3}} f (x) + C$
Put $x^{3} = t ∴ 3 x^{2} d x = d t$
$\Rightarrow \int x^{3} e^{- 4 x^{3}} x^{2} d x = \frac{1}{3} \int t e^{- 4 t} d t$
$= \frac{1}{3} [t \frac{e^{- 4 t}}{- 4} - \int \frac{e^{- 4 t}}{- 4} d t]$
$= - \frac{e^{- 4 t}}{48} [4 t + 1] + C$
$= - \frac{e^{- 4 x^{3}}}{48} [4 x^{3} + 1] + C$
$∴ f (x) = - 1 - 4 x^{3}$

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