What is the integral of (3x^2+2x-1)/(4e^{3x)} ?

The integral of (3x^2+2x-1)/(4e^{3x)} is 1/36 (-9e^{-3x}x^2-12e^{-3x}x-e^{-3x})+C

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integral of (3x^2+2x-1)/(4e^{3x)}

\int \frac{3 x ^{2} + 2 x - 1}{4 e ^{3 x}} d x

\frac{1}{36} (- 9 e^{- 3 x} x^{2} - 12 e^{- 3 x} x - e^{- 3 x}) + C

Solution steps

\int \frac{3 x ^{2} + 2 x - 1}{4 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}{4} \cdot \int \frac{3 x ^{2} + 2 x - 1}{e ^{3 x}} d x

Expand

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

= \frac{1}{4} \cdot \int \frac{3 x ^{2}}{e ^{3 x}} + \frac{2 x}{e ^{3 x}} - \frac{1}{e ^{3 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

= \frac{1}{4} (\int \frac{3 x ^{2}}{e ^{3 x}} d x + \int \frac{2 x}{e ^{3 x}} d x - \int \frac{1}{e ^{3 x}} d x)

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

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

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

= \frac{1}{4} (- e^{- 3 x} x^{2} - \frac{2}{3} e^{- 3 x} x - \frac{2}{9} e^{- 3 x} + 2 (- \frac{1}{3} e^{- 3 x} x - \frac{1}{9} e^{- 3 x}) - (- \frac{1}{3} e^{- 3 x}))

Simplify

\frac{1}{4} (- e^{- 3 x} x^{2} - \frac{2}{3} e^{- 3 x} x - \frac{2}{9} e^{- 3 x} + 2 (- \frac{1}{3} e^{- 3 x} x - \frac{1}{9} e^{- 3 x}) - (- \frac{1}{3} e^{- 3 x})) : \frac{1}{36} (- 9 e^{- 3 x} x^{2} - 12 e^{- 3 x} x - e^{- 3 x})

= \frac{1}{36} (- 9 e^{- 3 x} x^{2} - 12 e^{- 3 x} x - e^{- 3 x})

Add a constant to the solution

= \frac{1}{36} (- 9 e^{- 3 x} x^{2} - 12 e^{- 3 x} x - e^{- 3 x}) + C

\frac{d}{d x} (\frac{1 - x ^{2}}{5})

x^{2} - 2 x + 2

\int - \frac{x}{x ^{2} + 9} d x

arccos (4 x)

\int \frac{x ^{3}}{x ^{4} + 7} d x

What is the integral of (3x^2+2x-1)/(4e^{3x)} ?
The integral of (3x^2+2x-1)/(4e^{3x)} is 1/36 (-9e^{-3x}x^2-12e^{-3x}x-e^{-3x})+C