What is the integral of e^{-17x} ?

The integral of e^{-17x} is -1/17 e^{-17x}+C

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integral of e^{-17x}

\int e^{- 17 x} d x

- \frac{1}{17} e^{- 17 x} + C

Solution steps

\int e^{- 17 x} d x

Apply u-substitution

: \int - \frac{1}{17} e^{u} d u

= \int - \frac{1}{17} e^{u} d u

Take the constant out:

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

= - \frac{1}{17} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= - \frac{1}{17} e^{u}

Substitute back

u = - 17 x

= - \frac{1}{17} e^{- 17 x}

Add a constant to the solution

= - \frac{1}{17} e^{- 17 x} + C

\int \frac{x}{16 - x ^{8}} d x

\frac{d y}{d x} = \frac{2 x + y}{3 + 3 y ^{2} - x}, y (0) = 0

\frac{d}{d x} (ln (sin (x)))

(3, 5), (- 2, - 20)

(1 - x)^{2}