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

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

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

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

- e^{1 - x} + e x + C

Solution steps

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

Expand

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

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

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

= - e^{1 - x} + e x

Add a constant to the solution

= - e^{1 - x} + e x + C

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