What is the integral of (7x^2-8x)e^{2x} ?

The integral of (7x^2-8x)e^{2x} is 7/2 e^{2x}x^2-15/2 e^{2x}x+15/4 e^{2x}+C

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integral of (7x^2-8x)e^{2x}

\int (7 x^{2} - 8 x) e^{2 x} d x

\frac{7}{2} e^{2 x} x^{2} - \frac{15}{2} e^{2 x} x + \frac{15}{4} e^{2 x} + C

Solution steps

\int (7 x^{2} - 8 x) e^{2 x} d x

Expand

(7 x^{2} - 8 x) e^{2 x} : 7 e^{2 x} x^{2} - 8 e^{2 x} x

= \int 7 e^{2 x} x^{2} - 8 e^{2 x} 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 7 e^{2 x} x^{2} d x - \int 8 e^{2 x} x d x

\int 7 e^{2 x} x^{2} d x = 7 (\frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x}))

\int 8 e^{2 x} x d x = 2 (2 e^{2 x} x - e^{2 x})

= 7 (\frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})) - 2 (2 e^{2 x} x - e^{2 x})

Simplify

7 (\frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})) - 2 (2 e^{2 x} x - e^{2 x}) : \frac{7}{2} e^{2 x} x^{2} - \frac{15}{2} e^{2 x} x + \frac{15}{4} e^{2 x}

= \frac{7}{2} e^{2 x} x^{2} - \frac{15}{2} e^{2 x} x + \frac{15}{4} e^{2 x}

Add a constant to the solution

= \frac{7}{2} e^{2 x} x^{2} - \frac{15}{2} e^{2 x} x + \frac{15}{4} e^{2 x} + C

y^{^{'}} = \frac{- y}{x}

\frac{d}{d x} (\frac{x}{6 + x})

\frac{\partial}{\partial x} (\frac{1}{2 + x y ^{2}})

x \to 0 lim (\frac{x - x}{x})

\frac{d}{d x} ((x^{2} + 2)^{2})

What is the integral of (7x^2-8x)e^{2x} ?
The integral of (7x^2-8x)e^{2x} is 7/2 e^{2x}x^2-15/2 e^{2x}x+15/4 e^{2x}+C