What is the integral of (e^{x-5}-e^{3+x})/(e^{1-x)} ?

The integral of (e^{x-5}-e^{3+x})/(e^{1-x)} is (1-e^8)/2 e^{2x-6}+C

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integral of (e^{x-5}-e^{3+x})/(e^{1-x)}

\int \frac{e ^{x - 5} - e ^{3 + x}}{e ^{1 - x}} d x

\frac{1 - e ^{8}}{2} e^{2 x - 6} + C

Solution steps

\int \frac{e ^{x - 5} - e ^{3 + x}}{e ^{1 - x}} d x

Expand

\frac{e ^{x - 5} - e ^{3 + x}}{e ^{1 - x}} : \frac{e ^{x - 5}}{e ^{1 - x}} - \frac{e ^{3 + x}}{e ^{1 - 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 \frac{e ^{x - 5}}{e ^{1 - x}} d x - \int \frac{e ^{3 + x}}{e ^{1 - x}} d x

\int \frac{e ^{x - 5}}{e ^{1 - x}} d x = \frac{1}{2} e^{2 x - 6}

\int \frac{e ^{3 + x}}{e ^{1 - x}} d x = \frac{1}{2} e^{2 x + 2}

= \frac{1}{2} e^{2 x - 6} - \frac{1}{2} e^{2 x + 2}

Simplify

\frac{1}{2} e^{2 x - 6} - \frac{1}{2} e^{2 x + 2} : \frac{1 - e ^{8}}{2} e^{2 x - 6}

= \frac{1 - e ^{8}}{2} e^{2 x - 6}

Add a constant to the solution

= \frac{1 - e ^{8}}{2} e^{2 x - 6} + C

f (x) = (1 + 8 x^{2}) (x - x^{2})

\frac{d}{d x} (e^{x - 8})

\frac{d}{d x} (a^{x} + x^{n})

\frac{\partial}{\partial x} (6 x e^{x y})

\frac{\partial}{\partial x} (15 x^{2} + y^{2})

What is the integral of (e^{x-5}-e^{3+x})/(e^{1-x)} ?
The integral of (e^{x-5}-e^{3+x})/(e^{1-x)} is (1-e^8)/2 e^{2x-6}+C