The correct option is A e ^ 2x 4 ( 2 x ^ 4 4 x ^ 3 +6 x ^ 2 6x+3 ) +C ∫ x^4.e^2xdx Applying by parts rule of integration, x^42 e^2x 42∫ x^3.e^ ...

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Integration by Parts

∫ x 4 e 2x dx...

Question

$\int x^{4} e^{2 x} d x =$

\frac{e^{2 x}}{4} (2 x^{4} - 4 x^{3} + 6 x^{2} - 6 x + 3) + C

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\frac{e^{2 x}}{2} (2 x^{4} - 4 x^{3} + 6 x^{2} - 6 x + 3) + C

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\frac{e^{2 x}}{8} (2 x^{4} + 4 x^{3} + 6 x^{2} + 6 x + 3) + C

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\frac{e^{2 x}}{4} (2 x^{4} + 4 x^{3} + 6 x^{2} + 6 x + 3) + C

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Solution

The correct option is A $\frac{e^{2 x}}{4} (2 x^{4} - 4 x^{3} + 6 x^{2} - 6 x + 3) + C$
$\int x^{4} . e^{2 x} d x$

Applying by parts rule of integration,

$\frac{x^{4}}{2} e^{2 x} - \frac{4}{2} \int x^{3} . e^{2 x} d x$

Applying by-parts again, multiple times till the time powers in $x$ in the integral vanishes,

$\frac{x^{4}}{2} e^{2 x} - 2 [\frac{x^{3}}{2} e^{2 x} - \frac{3}{2} \int x^{2} . e^{2 x} d x]$
=> $\frac{x^{4}}{2} e^{2 x} - x^{3} . e^{2 x} + 3 [\frac{x^{2}}{2} . e^{2 x} - \int x . e^{2 x} d x]$
=> $\frac{x^{4}}{2} e^{2 x} - x^{3} . e^{2 x} + 3 \frac{x^{2}}{2} . e^{2 x} - 3 [\frac{x}{2} . e^{2 x} - \int \frac{e^{2 x}}{2} d x]$
=> $\frac{x^{4}}{2} e^{2 x} - x^{3} . e^{2 x} + 3 \frac{x^{2}}{2} . e^{2 x} - 3 \frac{x}{2} . e^{2 x} + 3 \frac{e^{2 x}}{4} + c$
$\to \frac{e^{2 x}}{4} (2 x^{4} - 4 x^{3} + 6 x^{2} - 6 x + 3) + C$

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