int(x^(2)+2x+100)e^(-3x)dx=

Identify integral: Identify the integral to be solved. We need to find the integral of the function (x^2 + 2x + 100)e^(-3x) with respect to x.Apply integration by parts: Apply integration by parts. Integration by parts formula is ∫u dv = uv - ∫v du. Let u = x^2 + 2x + 100, which means du = (2x + 2)dx. Let dv = e^(-3x)dx, which means v = -1/3 e^(-3x).Perform first integration: Perform the first integration by parts. Using the formula, we get: ∫(x^2 + 2x + 100)e^(-3x)dx = (x^2 + 2x + 100)(-1/3 e^(-3x)) - ∫(-1/3 e^(-3x))(2x + 2)dxSimplify expression: Simplify the expression. = -(1/3)(x^2 + 2x + 100)e^(-3x) + (2/3)∫(x + 1)e^(-3x)dx Now we need to integrate (x + 1)e^(-3x) with respect to x.Apply integration by parts again: Apply integration by parts again for the remaining integral. Let u = x + 1, which means du = dx. Let dv = e^(-3x)dx, which means v = -1/3 e^(-3x).Perform second integration: Perform the second integration by parts. ∫(x + 1)e^(-3x)dx = (x + 1)(-1/3 e^(-3x)) - ∫(-1/3 e^(-3x))(1)dxIntegrate e^(-3x): Simplify the expression. = -(1/3)(x + 1)e^(-3x) + (1/3)∫e^(-3x)dx Now we need to integrate e^(-3x) with respect to x.Combine all parts: Integrate e^(-3x) with respect to x. ∫e^(-3x)dx = -1/3 e^(-3x) + C, where C is the constant of integration.Simplify final expression: Combine all parts together. Putting it all together, we have: ∫(x^2 + 2x + 100)e^(-3x)dx = -(1/3)(x^2 + 2x + 100)e^(-3x) + (2/3)(-(1/3)(x + 1)e^(-3x) + (1/3)∫e^(-3x)dx) = -(1/3)(x^2 + 2x + 100)e^(-3x) - (2/9)(x + 1)e^(-3x) - (2/9)(-1/3 e^(-3x)) + CSimplify the final expression. = -(1/3)(x^2 + 2x + 100)e^(-3x) - (2/9)(x + 1)e^(-3x) + (2/27)e^(-3x) + C = -e^(-3x)(1/3 x^2 + 2/3 x + 100/3 + 2/9 x + 2/9 - 2/27) + C = -e^(-3x)(1/3 x^2 + 8/9 x + 298/27) + C

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$\int\left(x^{2}+2 x+100\right) e^{-3 x} d x$ =

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\int\left(x^{2}+2 x+100\right) e^{-3 x} d x

Identify integral: Identify the integral to be solved. $\newline$ We need to find the integral of the function $(x^2 + 2x + 100)e^{-3x}$ with respect to $x$ .
Apply integration by parts: Apply integration by parts. $\newline$ Integration by parts formula is $\int u \, dv = uv - \int v \, du$ . $\newline$ Let $u = x^2 + 2x + 100$ , which means $du = (2x + 2)dx$ . $\newline$ Let $dv = e^{-3x}dx$ , which means $v = -\frac{1}{3} e^{-3x}$ .
Perform first integration: Perform the first integration by parts. $\newline$ Using the formula, we get: $\newline$ $\int(x^2 + 2x + 100)e^{-3x}\,dx = (x^2 + 2x + 100)(-\frac{1}{3} e^{-3x}) - \int(-\frac{1}{3} e^{-3x})(2x + 2)\,dx$
Simplify expression: Simplify the expression. $\newline$ $= -(\frac{1}{3})(x^2 + 2x + 100)e^{-3x} + (\frac{2}{3})\int(x + 1)e^{-3x}dx$ $\newline$ Now we need to integrate $(x + 1)e^{-3x}$ with respect to $x$ .
Apply integration by parts again: Apply integration by parts again for the remaining integral. $\newline$ Let $u = x + 1$ , which means $du = dx$ . $\newline$ Let $dv = e^{-3x}dx$ , which means $v = -\frac{1}{3} e^{-3x}$ .
Perform second integration: Perform the second integration by parts. $\newline$ $\int(x + 1)e^{-3x}dx = (x + 1)(-\frac{1}{3} e^{-3x}) - \int(-\frac{1}{3} e^{-3x})(1)dx$
Integrate $e^{-3x}$ : Simplify the expression. $\newline$ $= -\frac{1}{3}(x + 1)e^{-3x} + \frac{1}{3}\int e^{-3x}\,dx$ $\newline$ Now we need to integrate $e^{-3x}$ with respect to $x$ .
Combine all parts: Integrate $e^{-3x}$ with respect to $x$ .
$\int e^{-3x}\,dx = -\frac{1}{3} e^{-3x} + C$ , where $C$ is the constant of integration.
Simplify final expression: Combine all parts together. $\newline$ Putting it all together, we have: $\newline$ $\int(x^2 + 2x + 100)e^{(-3x)}dx = -(\frac{1}{3})(x^2 + 2x + 100)e^{(-3x)} + (\frac{2}{3})(-(\frac{1}{3})(x + 1)e^{(-3x)} + (\frac{1}{3})\int e^{(-3x)}dx)$ $\newline$ $= -(\frac{1}{3})(x^2 + 2x + 100)e^{(-3x)} - (\frac{2}{9})(x + 1)e^{(-3x)} - (\frac{2}{9})(-\frac{1}{3} e^{(-3x)}) + C$
Simplify final expression: Combine all parts together. $\newline$ Putting it all together, we have: $\newline$ $\int(x^2 + 2x + 100)e^{-3x}\,dx = -(\frac{1}{3})(x^2 + 2x + 100)e^{-3x} + (\frac{2}{3})(-(\frac{1}{3})(x + 1)e^{-3x} + (\frac{1}{3})\int e^{-3x}\,dx)$ $\newline$ $= -(\frac{1}{3})(x^2 + 2x + 100)e^{-3x} - (\frac{2}{9})(x + 1)e^{-3x} - (\frac{2}{9})(-\frac{1}{3} e^{-3x}) + C$ Simplify the final expression. $\newline$ $= -(\frac{1}{3})(x^2 + 2x + 100)e^{-3x} - (\frac{2}{9})(x + 1)e^{-3x} + (\frac{2}{27})e^{-3x} + C$ $\newline$ $= -e^{-3x}(\frac{1}{3} x^2 + \frac{2}{3} x + \frac{100}{3} + \frac{2}{9} x + \frac{2}{9} - \frac{2}{27}) + C$ $\newline$ $= -e^{-3x}(\frac{1}{3} x^2 + \frac{8}{9} x + \frac{298}{27}) + C$

$\int\left(x^{2}+2 x+100\right) e^{-3 x} d x$ =

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∫(x2+2x+100)e−3xdx \int\left(x^{2}+2 x+100\right) e^{-3 x} d x ∫(x2+2x+100)e−3xdx=

$\int\left(x^{2}+2 x+100\right) e^{-3 x} d x$ =