What is the integral of ((e^{2x}+1))/(e^x) ?

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

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integral of ((e^{2x}+1))/(e^x)

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

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

Solution steps

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

\frac{e ^{2 x} + 1}{e ^{x}} = (\frac{e ^{2 x}}{e ^{x}} + \frac{1}{e ^{x}})

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

\int \frac{e ^{2 x}}{e ^{x}} d x = e^{x}

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

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

Add a constant to the solution

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

y^{^{'}} = \frac{1}{y}, y (0) = 1

f (x) = 3 x^{2} + 1

\frac{d}{d x} (ln^{3} (2 x + 5))

x y^{^{'}} + y = x^{4} y^{3}

\frac{d}{d x} ((6 x - 7)^{3} (8 x^{2} + 9)^{2})

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