What is the integral of (x^2)/(9e^{3x)} ?

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

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integral of (x^2)/(9e^{3x)}

\int \frac{x ^{2}}{9 e ^{3 x}} d x

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

Solution steps

\int \frac{x ^{2}}{9 e ^{3 x}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{9} \cdot \int \frac{x ^{2}}{e ^{3 x}} d x

Apply Integration By Parts

: - \frac{1}{3} e^{- 3 x} x^{2} - \int - \frac{2}{3} e^{- 3 x} x d x

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

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

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

Simplify

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

Add a constant to the solution

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

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