What is the integral of (x^2e^{2x}-3xe^{2x}+5e^{2x}) ?

The integral of (x^2e^{2x}-3xe^{2x}+5e^{2x}) is 1/2 e^{2x}x^2-2e^{2x}x+7/2 e^{2x}+C

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integral of (x^2e^{2x}-3xe^{2x}+5e^{2x})

\int (x^{2} e^{2 x} - 3 x e^{2 x} + 5 e^{2 x}) d x

\frac{1}{2} e^{2 x} x^{2} - 2 e^{2 x} x + \frac{7}{2} e^{2 x} + C

Solution steps

\int x^{2} e^{2 x} - 3 x e^{2 x} + 5 e^{2 x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{2} e^{2 x} d x - \int 3 x e^{2 x} d x + \int 5 e^{2 x} d x

\int x^{2} e^{2 x} d x = \frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})

\int 3 x e^{2 x} d x = \frac{3}{4} (2 e^{2 x} x - e^{2 x})

\int 5 e^{2 x} d x = \frac{5}{2} e^{2 x}

= \frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x}) - \frac{3}{4} (2 e^{2 x} x - e^{2 x}) + \frac{5}{2} e^{2 x}

Simplify

\frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x}) - \frac{3}{4} (2 e^{2 x} x - e^{2 x}) + \frac{5}{2} e^{2 x} : \frac{1}{2} e^{2 x} x^{2} - 2 e^{2 x} x + \frac{7}{2} e^{2 x}

= \frac{1}{2} e^{2 x} x^{2} - 2 e^{2 x} x + \frac{7}{2} e^{2 x}

Add a constant to the solution

= \frac{1}{2} e^{2 x} x^{2} - 2 e^{2 x} x + \frac{7}{2} e^{2 x} + C

\int_{2}^{6} 8 d x

x \to 0 lim (3 sin (x) - 1)

\frac{2 x + 5}{x}

\frac{d}{d x} (\frac{x ^{3}}{f ( x )})

\int \frac{9}{t ^{2}} d t

What is the integral of (x^2e^{2x}-3xe^{2x}+5e^{2x}) ?
The integral of (x^2e^{2x}-3xe^{2x}+5e^{2x}) is 1/2 e^{2x}x^2-2e^{2x}x+7/2 e^{2x}+C