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

The integral of (2x^2+x)e^{3x} is 2/3 e^{3x}x^2-1/9 e^{3x}x+1/27 e^{3x}+C

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

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

\frac{2}{3} e^{3 x} x^{2} - \frac{1}{9} e^{3 x} x + \frac{1}{27} e^{3 x} + C

Solution steps

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

Expand

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

= \int 2 e^{3 x} x^{2} + 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 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 e^{3 x} x d x = \frac{1}{9} (e^{3 x} \cdot 3 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{1}{9} (e^{3 x} \cdot 3 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{1}{9} (e^{3 x} \cdot 3 x - e^{3 x}) : \frac{2}{3} e^{3 x} x^{2} - \frac{1}{9} e^{3 x} x + \frac{1}{27} e^{3 x}

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

Add a constant to the solution

= \frac{2}{3} e^{3 x} x^{2} - \frac{1}{9} e^{3 x} x + \frac{1}{27} e^{3 x} + C

x \to 0 lim (\frac{x ^{2} - 3 x}{x})

f (x) = \frac{1}{( x - 1 ) ^{2}}, at x = 0

\int 2 π x d x

y = x^{2} - 3

y^{^{'}} e^{- x} = y^{2} + 9

What is the integral of (2x^2+x)e^{3x} ?
The integral of (2x^2+x)e^{3x} is 2/3 e^{3x}x^2-1/9 e^{3x}x+1/27 e^{3x}+C