What is the integral of (e^{ax}-e^{bx})/(e^{cx)} ?

The integral of (e^{ax}-e^{bx})/(e^{cx)} is 1/(a-c)e^{ax-cx}-1/(b-c)e^{bx-cx}+C

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integral of (e^{ax}-e^{bx})/(e^{cx)}

\int \frac{e ^{a x} - e ^{b x}}{e ^{c x}} d x

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

Solution steps

\int \frac{e ^{a x} - e ^{b x}}{e ^{c x}} d x

Expand

\frac{e ^{a x} - e ^{b x}}{e ^{c x}} : e^{a x - c x} - e^{b x - c x}

= \int e^{a x - c x} - e^{b x - c x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int e^{a x - c x} d x - \int e^{b x - c x} d x

\int e^{a x - c x} d x = \frac{1}{a - c} e^{a x - c x}

\int e^{b x - c x} d x = \frac{1}{b - c} e^{b x - c x}

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

Add a constant to the solution

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

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\int \frac{x ^{2} + 2 x + 1}{( x - 2 ) ^{2} ( x - 1 )} d x

\int x (x^{2} + 5)^{9} d x

\int \frac{cos ( x )}{sin ^{2} ( x ) + sin ( x ) - 2} d x

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