# Use the reduction formulas in a table of integrals to evaluate the following integrals. int x^3e^{2x}dx

Use the reduction formulas in a table of integrals to evaluate the following integrals.
$\int {x}^{3}{e}^{2x}dx$
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izboknil3

$\text{For evaluating given integral, we use integral by parts theorem}$
$\text{According to integral by parts theorem}$
$\int f\left(x\right){g}^{\prime }\left(x\right)dx=f\left(x\right)\int {g}^{\prime }\left(x\right)dx-\left[\int {f}^{\prime }\left(x\right)\int {g}^{\prime }\left(x\right)dx\right)dx\right]$

$\text{So, be using equation}$
$I=\int {x}^{3}{e}^{2x}dx$
$={x}^{3}\int {e}^{2x}dx-\left[\int \left(\frac{d}{dx}\left({x}^{3}\right)\int {e}^{2x}dx\right)dx\right]$
$\left(\because \int {e}^{kx}dx=\frac{{e}^{kx}}{k}+c,\frac{d}{dx}\left({x}^{n}\right)=\left(n{x}^{n-1}\right)$
$={x}^{3}\left(\frac{{e}^{2x}}{2}\right)-\left[\int \left(3{x}^{2}\right)\left(\frac{{e}^{2x}}{2}dx\right]$
$=\frac{{x}^{3}{e}^{2x}}{2}-\frac{3}{2}\left[\int {x}^{2}{e}^{2x}dx\right]$
$=\frac{{x}^{3}{e}^{2x}}{2}-\frac{3}{2}\left[{x}^{2}\int {e}^{2x}dx-\left[\int \left[\frac{d}{dx}\left({x}^{2}\right)\int {e}^{2x}dx\right]dx\right]\right]$
$=\frac{{x}^{3}{e}^{2x}}{2}-\frac{3}{2}\left[{x}^{2}\left(\frac{{e}^{2x}}{2}\right)-\left[\int \left(2x\right)\frac{{e}^{2x}}{2}dx\right]\right]$
$=\frac{{x}^{3}{e}^{2x}}{2}-\frac{3{x}^{2}{e}^{2x}}{4}+\frac{3}{2}\int x{e}^{2x}dx$
$=\frac{{x}^{3}{e}^{2x}}{2}-\frac{3{x}^{2}{e}^{2x}}{4}+\frac{3}{2}\left[x\int {e}^{2x}dx-\left[\int \left(\frac{d}{dx}\left(x\right)\int {e}^{2x}dx\right)dx\right]\right]$
$=\frac{{x}^{3}{e}^{2x}}{2}-\frac{3{x}^{2}{e}^{2x}}{4}+\frac{3}{2}\left[x\left(\frac{{e}^{2x}}{2}-\left[\int \left(1\right)\left(\frac{{e}^{2x}}{2}\right)dx\right]\right]$
$=\frac{{x}^{3}{e}^{2x}}{2}-\frac{3{x}^{2}{e}^{2x}}{4}+\frac{3x{e}^{2x}}{4}-\frac{3{e}^{2x}}{8}+c$
$=\frac{\left(4{x}^{3}-6{x}^{2}+6x-3\right){e}^{2x}}{8}+c$