What is the integral of ((e^{4x}-1)/(e^{3x)}) ?

The integral of ((e^{4x}-1)/(e^{3x)}) is e^x+1/3 e^{-3x}+C

integral of ((e^{4x}-1)/(e^{3x)})

\int (\frac{e ^{4 x} - 1}{e ^{3 x}}) d x

e^{x} + \frac{1}{3} e^{- 3 x} + C

Solution steps

\int \frac{e ^{4 x} - 1}{e ^{3 x}} d x

Expand

\frac{e ^{4 x} - 1}{e ^{3 x}} : e^{x} - \frac{1}{e ^{3 x}}

= \int e^{x} - \frac{1}{e ^{3 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^{x} d x - \int \frac{1}{e ^{3 x}} d x

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

\int \frac{1}{e ^{3 x}} d x = - \frac{1}{3} e^{- 3 x}

= e^{x} - (- \frac{1}{3} e^{- 3 x})

Simplify

= e^{x} + \frac{1}{3} e^{- 3 x}

Add a constant to the solution

= e^{x} + \frac{1}{3} e^{- 3 x} + C

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What is the integral of ((e^{4x}-1)/(e^{3x)}) ?
The integral of ((e^{4x}-1)/(e^{3x)}) is e^x+1/3 e^{-3x}+C