Answer:
Option A
Explanation:
Let I= $\int x^{4}e^{2x} dx$
$\Rightarrow I=x^{4}\int e^{2x} dx-\int \frac{dx^{4}}{dx}.\int e^{2x} .dx.dx+C$
$=\frac{x^{4}.e^{2x}}{2}-\frac{1}{2}\int 4x^{3}e^{2x}dx+C$
$\Rightarrow\frac{x^{4}.e^{2x}}{2}-2\left[\frac{x^{3}e^{2x}}{2}-\int\frac{3x^{2}e^{2x}}{2}dx\right]+C$
$\Rightarrow\frac{x^{4}.e^{2x}}{2}-x^{3}e^{2x}+3\left[\frac{x^{2}e^{2x}}{2}-\frac{1}{2}\int2xe^{2x} dx\right]+C$
$\Rightarrow\frac{x^{4}.e^{2x}}{2}-x^{3}e^{2x}+\frac{3}{2} x^{2}e^{2x}-3$
$\left[\frac{xe^{2x}}{2}-\int\frac{e^{2x}}{2}dx\right]+C $
$\Rightarrow\frac{x^{4}.e^{2x}}{2}-x^{3}e^{2x}+\frac{3}{2} x^{2}e^{2x}-\frac{3}{2}xe^{2x}+\frac{3e^{2x}}{4}+C$
$\Rightarrow\frac{e^{2x}}{4}\left[ 2x^{4}-4x^{3}+6x^{2}-6x+3\right]+C$
