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

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

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integral of 17e^{-17x}

\int 17 e^{- 17 x} d x

- e^{- 17 x} + C

Solution steps

\int 17 e^{- 17 x} d x

Take the constant out:

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

= 17 \cdot \int e^{- 17 x} d x

Apply u-substitution

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

= 17 \cdot \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

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

Use the common integral:

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

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

Substitute back

u = - 17 x

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

Simplify

17 (- \frac{1}{17} e^{- 17 x}) : - e^{- 17 x}

= - e^{- 17 x}

Add a constant to the solution

= - e^{- 17 x} + C

x \to 0 lim (x (1 + \frac{1}{x}))

y = \frac{( x - 1 )}{( x + 1 )}

\frac{\partial}{\partial y} (ln (\frac{x y}{y ^{2} + z ^{2}}))

\frac{\partial}{\partial y} (y e^{y x})

3 cos (x) - 2 sin (x)