What is the integral of (e^{2y}-y)/(e^y) ?

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

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integral of (e^{2y}-y)/(e^y)

\int \frac{e ^{2 y} - y}{e ^{y}} d y

e^{y} + \frac{y}{e ^{y}} + \frac{1}{e ^{y}} + C

Solution steps

\int \frac{e ^{2 y} - y}{e ^{y}} d y

Expand

\frac{e ^{2 y} - y}{e ^{y}} : e^{y} - \frac{y}{e ^{y}}

= \int e^{y} - \frac{y}{e ^{y}} d y

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^{y} d y - \int \frac{y}{e ^{y}} d y

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

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

= e^{y} - (- \frac{y}{e ^{y}} - \frac{1}{e ^{y}})

Simplify

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

Add a constant to the solution

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

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