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

The integral of (e^{-3x}-1)^2 is -1/6 e^{-6x}+2/3 e^{-3x}+x+C

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

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

- \frac{1}{6} e^{- 6 x} + \frac{2}{3} e^{- 3 x} + x + C

Solution steps

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

Expand

(e^{- 3 x} - 1)^{2} : e^{- 6 x} - 2 e^{- 3 x} + 1

= \int e^{- 6 x} - 2 e^{- 3 x} + 1 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 e^{- 6 x} d x - \int 2 e^{- 3 x} d x + \int 1 d x

\int e^{- 6 x} d x = - \frac{1}{6} e^{- 6 x}

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

\int 1 d x = x

= - \frac{1}{6} e^{- 6 x} - (- \frac{2}{3} e^{- 3 x}) + x

Simplify

= - \frac{1}{6} e^{- 6 x} + \frac{2}{3} e^{- 3 x} + x

Add a constant to the solution

= - \frac{1}{6} e^{- 6 x} + \frac{2}{3} e^{- 3 x} + x + C

\frac{d y}{d x} - 2 y = 2 x

\frac{d}{d x} (x^{\frac{8}{5}})

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

\frac{d}{d θ} (sec (θ) \cdot tan (θ))

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

What is the integral of (e^{-3x}-1)^2 ?
The integral of (e^{-3x}-1)^2 is -1/6 e^{-6x}+2/3 e^{-3x}+x+C