$\int e^{x} 2^{3 {log}_{2} x} d x = e^{x} f (x) + c$ , then $f (x) =$

x^{3} - 3 x^{2} + 6 x - 6

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x^{3} - 3 x^{2} - 6 x - 3

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x^{3} - 3 x^{2} + 6 x + 6

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x^{3} + 3 x^{2} + 6 x + 6

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Solution

The correct option is B $x^{3} - 3 x^{2} + 6 x - 6$
Given, $\int e^{x} 2^{3 {log}_{2} x} d x = e^{x} f (x) + c$ ......(A)
Consider, $\int e^{x} 2^{3 {log}_{2} x} d x = \int e^{x} . x^{3} d x$
$\int e^{x} \times x^{3} d x = x^{3} \int e^{x} d x - \int 3 x^{2} e^{x} d x$
$= x^{3} e^{x} - 3 \int x^{2} e^{x} d x$ ....(1)
$\int x^{2} e^{x} = x^{2} e^{x} - 2 \int x e^{x} d x$ .....(2)
$\int x e^{x} = x e^{x} - e^{x}$ .....(3)
Using (3) in (2),
$\int x^{2} e^{x} = x^{2} e^{x} - 2 (x e^{x} - e^{x})$ ....(4)
Using (4) in (1)
$\int e^{x} . x^{3} d x = x^{3} e^{x} - 3 [x^{2} e^{x} - 2 (x e^{x} - e^{x})] + c$
$= x^{3} e^{x} - 3 x^{2} e^{x} + 6 x e^{x} - 6 e^{x} + c$
$= e^{x} [x^{3} - 3 x^{2} + 6 x - 6] + c$
On comparing with eqn (A), we get
$f (x) = x^{3} - 3 x^{2} + 6 x - 6$

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