int x^4 e^{2 x} d x= | vedclass

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(A) Let $I = \int x^4 e^{2 x} d x$.Using integration by parts, $\int u v d x = u \int v d x - \int (u' \int v d x) d x$.Taking $u = x^4$ and $v = e^{2

Vedclass · Accepted Answer

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

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(A) Let $I = \int x^4 e^{2 x} d x$.Using integration by parts, $\int u v d x = u \int v d x - \int (u' \int v d x) d x$.Taking $u = x^4$ and $v = e^{2 x}$:$I = x^4 \frac{e^{2 x}}{2} - \int 4 x^3 \frac{e^{2 x}}{2} d x = \frac{x^4 e^{2 x}}{2} - 2 \int x^3 e^{2 x} d x$.Applying integration by parts again:$\int x^3 e^{2 x} d x = x^3 \frac{e^{2 x}}{2} - \int 3 x^2 \frac{e^{2 x}}{2} d x = \frac{x^3 e^{2 x}}{2} - \frac{3}{2} \int x^2 e^{2 x} d x$.Continuing the process:$\int x^2 e^{2 x} d x = \frac{x^2 e^{2 x}}{2} - \int x e^{2 x} d x = \frac{x^2 e^{2 x}}{2} - (\frac{x e^{2 x}}{2} - \frac{e^{2 x}}{4})$.Substituting back into the expression for $I$:$I = \frac{x^4 e^{2 x}}{2} - 2 [\frac{x^3 e^{2 x}}{2} - \frac{3}{2} (\frac{x^2 e^{2 x}}{2} - \frac{x e^{2 x}}{2} + \frac{e^{2 x}}{4})] + C$.$I = \frac{x^4 e^{2 x}}{2} - x^3 e^{2 x} + \frac{3}{2} x^2 e^{2 x} - \frac{3}{2} x e^{2 x} + \frac{3}{4} e^{2 x} + C$.$I = \frac{e^{2 x}}{4} (2 x^4 - 4 x^3 + 6 x^2 - 6 x + 3) + C$.

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

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