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

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

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

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

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

Solution steps

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

Expand

\frac{e ^{2 x} + 2 e ^{x} + 1}{e ^{x}} : e^{x} + 2 + \frac{1}{e ^{x}}

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

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

\int 2 d x = 2 x

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

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

Add a constant to the solution

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

\frac{d}{d x} (csc (4 x))

y = 5 x - x^{2}, y = 2 x

\int \frac{5 x x}{1 + x ^{5}} d x

y^{^{''}} - 18 y^{^{'}} + 145 y = 0

8 sin (x) + 6 cos (x), at x = 0

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