What is the integral of (5x^2-8x)e^{2x} ?

The integral of (5x^2-8x)e^{2x} is 5/2 e^{2x}x^2-13/2 e^{2x}x+13/4 e^{2x}+C

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integral of (5x^2-8x)e^{2x}

\int (5 x^{2} - 8 x) e^{2 x} d x

\frac{5}{2} e^{2 x} x^{2} - \frac{13}{2} e^{2 x} x + \frac{13}{4} e^{2 x} + C

Solution steps

\int (5 x^{2} - 8 x) e^{2 x} d x

Expand

(5 x^{2} - 8 x) e^{2 x} : 5 e^{2 x} x^{2} - 8 e^{2 x} x

= \int 5 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 5 e^{2 x} x^{2} d x - \int 8 e^{2 x} x d x

\int 5 e^{2 x} x^{2} d x = 5 (\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})

= 5 (\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

5 (\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{5}{2} e^{2 x} x^{2} - \frac{13}{2} e^{2 x} x + \frac{13}{4} e^{2 x}

= \frac{5}{2} e^{2 x} x^{2} - \frac{13}{2} e^{2 x} x + \frac{13}{4} e^{2 x}

Add a constant to the solution

= \frac{5}{2} e^{2 x} x^{2} - \frac{13}{2} e^{2 x} x + \frac{13}{4} e^{2 x} + C

\int_{0}^{1} \frac{13}{x ^{5}} d x

2 x^{2}

f (x) = \frac{3}{5} x^{10}

\frac{d}{d x} \frac{( x ^{2} - 4 ) ^{3}}{6}

\int \frac{1}{y ( y + 1 )} d y

What is the integral of (5x^2-8x)e^{2x} ?
The integral of (5x^2-8x)e^{2x} is 5/2 e^{2x}x^2-13/2 e^{2x}x+13/4 e^{2x}+C