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}})

Simplify

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

Add a constant to the solution

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

x \to 3 lim (2 x)

y^{^{'}} = cos (x - y)

\frac{9 e ^{x} x ^{\frac{7}{2}}}{2} + e^{x} x^{\frac{9}{2}}

e^{x}, e^{- 2 x}, ln (4)

\int_{0}^{3} x^{3} - x^{2} - 6 x d x

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