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

The integral of (2x^2+5x)e^{3x} is 2/3 e^{3x}x^2+11/9 e^{3x}x-11/27 e^{3x}+C

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

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

\frac{2}{3} e^{3 x} x^{2} + \frac{11}{9} e^{3 x} x - \frac{11}{27} e^{3 x} + C

Solution steps

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

Expand

(2 x^{2} + 5 x) e^{3 x} : 2 e^{3 x} x^{2} + 5 e^{3 x} x

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

\int 2 e^{3 x} x^{2} d x = 2 (\frac{1}{3} e^{3 x} x^{2} - \frac{2}{27} (3 e^{3 x} x - e^{3 x}))

\int 5 e^{3 x} x d x = \frac{5}{9} (3 e^{3 x} x - e^{3 x})

= 2 (\frac{1}{3} e^{3 x} x^{2} - \frac{2}{27} (3 e^{3 x} x - e^{3 x})) + \frac{5}{9} (3 e^{3 x} x - e^{3 x})

Simplify

2 (\frac{1}{3} e^{3 x} x^{2} - \frac{2}{27} (3 e^{3 x} x - e^{3 x})) + \frac{5}{9} (3 e^{3 x} x - e^{3 x}) : \frac{2}{3} e^{3 x} x^{2} + \frac{11}{9} e^{3 x} x - \frac{11}{27} e^{3 x}

= \frac{2}{3} e^{3 x} x^{2} + \frac{11}{9} e^{3 x} x - \frac{11}{27} e^{3 x}

Add a constant to the solution

= \frac{2}{3} e^{3 x} x^{2} + \frac{11}{9} e^{3 x} x - \frac{11}{27} e^{3 x} + C

\frac{e ^{- a s}}{s ^{2}}

3 + 4 x^{2} - 5 x^{3}

y^{^{''}} + 6 y^{^{'}} + 9 y = 4 e^{- 3 x}

\int \frac{1}{1 + \frac{x}{2}} d x

n = 1 \sum \infty \frac{3 ^{n}}{4 ^{n + 1}}

What is the integral of (2x^2+5x)e^{3x} ?
The integral of (2x^2+5x)e^{3x} is 2/3 e^{3x}x^2+11/9 e^{3x}x-11/27 e^{3x}+C