What is the integral of 1/(e^{kx)} ?

The integral of 1/(e^{kx)} is x/(e^{kx)}-e^{-kx}x-(e^{-kx})/k+C

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integral of 1/(e^{kx)}

\int \frac{1}{e ^{k x}} d x

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

Solution steps

\int \frac{1}{e ^{k x}} d x

Apply Integration By Parts

: \frac{x}{e ^{k x}} - \int - k e^{- k x} x d x

= \frac{x}{e ^{k x}} - \int - k e^{- k x} x d x

\int - k e^{- k x} x d x = e^{- k x} x + \frac{e ^{- k x}}{k}

= \frac{x}{e ^{k x}} - (e^{- k x} x + \frac{e ^{- k x}}{k})

Simplify

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

Add a constant to the solution

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

y^{^{''}} + y = 0, y (0) = 0, y (π) = 0

y = 2 x + 1 (4.3)

\int x (x - 1)^{5} d x

\frac{\partial}{\partial x} (y^{\frac{1}{3}} x^{\frac{2}{3}})

\int \frac{1}{( x - 1 ) ( x + 1 )} d x

What is the integral of 1/(e^{kx)} ?
The integral of 1/(e^{kx)} is x/(e^{kx)}-e^{-kx}x-(e^{-kx})/k+C