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

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

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

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

e^{x} + e x + C

Solution steps

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

Expand

e^{x} (1 + e^{1 - x}) : e^{x} + e

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

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

\int e d x = e x

= e^{x} + e x

Add a constant to the solution

= e^{x} + e x + C

x y^{^{'}} + 6 x^{2} y = x e^{5 x}

\int sec (x) tan^{2} (x) d x

\int \frac{( x - 2 )}{( x ^{2} - 4 x + 3 ) ^{3}} d x

\int \frac{x}{x ^{2} + 4 x - 5} d x

\frac{12}{s}

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