What is the integral of (2x-9)e^{2x} ?

The integral of (2x-9)e^{2x} is e^{2x}x-5e^{2x}+C

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integral of (2x-9)e^{2x}

\int (2 x - 9) e^{2 x} d x

e^{2 x} x - 5 e^{2 x} + C

Solution steps

\int (2 x - 9) e^{2 x} d x

Expand

(2 x - 9) e^{2 x} : 2 e^{2 x} x - 9 e^{2 x}

= \int 2 e^{2 x} x - 9 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 2 e^{2 x} x d x - \int 9 e^{2 x} d x

\int 2 e^{2 x} x d x = e^{2 x} x - \frac{1}{2} e^{2 x}

\int 9 e^{2 x} d x = \frac{9}{2} e^{2 x}

= e^{2 x} x - \frac{1}{2} e^{2 x} - \frac{9}{2} e^{2 x}

Add similar elements:

- \frac{1}{2} e^{2 x} - \frac{9}{2} e^{2 x} = - 5 e^{2 x}

= e^{2 x} x - 5 e^{2 x}

Add a constant to the solution

= e^{2 x} x - 5 e^{2 x} + C

\frac{d}{d x} (- \frac{3 y}{3 x + 2 y})

\frac{d}{d x} (\frac{1}{( 3 - x ) ^{2}})

cos (k \cdot t)

\frac{d y}{d x} = y^{\frac{1}{3}}, y (1) = 0

\frac{( x ^{2} + 1 ) d y}{d x} = - y^{2} - 1

What is the integral of (2x-9)e^{2x} ?
The integral of (2x-9)e^{2x} is e^{2x}x-5e^{2x}+C