What is the integral of e^{-kNx} ?

The integral of e^{-kNx} is -1/(kN)e^{-kNx}+C

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integral of e^{-kNx}

\int e^{- k N x} d x

- \frac{1}{k N} e^{- k N x} + C

Solution steps

\int e^{- k N x} d x

Apply u-substitution

: \int - \frac{e ^{u}}{k N} d u

= \int - \frac{e ^{u}}{k N} d u

Take the constant out:

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

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

Use the common integral:

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

= - \frac{1}{k N} e^{u}

Substitute back

u = - k N x

= - \frac{1}{k N} e^{- k N x}

Add a constant to the solution

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

y = A x^{2} + B e^{2 x}

y^{^{'}} = 2 x y, y (0) = 1

(x y) d x + (x^{2} + y^{2}) d y = 0

2 (x - 6)^{3}

What is the integral of e^{-kNx} ?
The integral of e^{-kNx} is -1/(kN)e^{-kNx}+C