What is the integral of e^{(a-b)x} ?

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

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integral of e^{(a-b)x}

\int e^{(a - b) x} d x

\frac{1}{a - b} e^{(a - b) x} + C

Solution steps

\int e^{(a - b) x} d x

Apply u-substitution

: \int \frac{e ^{u}}{a - b} d u

= \int \frac{e ^{u}}{a - b} d u

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{a - b} e^{u}

Substitute back

u = (a - b) x

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

Add a constant to the solution

= \frac{1}{a - b} e^{(a - b) x} + C

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

\int \frac{1}{20 - x ^{2}} d x

\int \frac{2 ln ^{2} ( x ) + 16}{x} d x

\frac{d}{d y} (2)

\frac{\partial}{\partial x} (n y ln (n x))

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