What is the integral of ((3+e^x)^2)/(e^x) ?

The integral of ((3+e^x)^2)/(e^x) is -9/(e^x)+6x+e^x+C

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integral of ((3+e^x)^2)/(e^x)

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

- \frac{9}{e ^{x}} + 6 x + e^{x} + C

Solution steps

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

Expand

\frac{( 3 + e ^{x} ) ^{2}}{e ^{x}} : \frac{9}{e ^{x}} + 6 + e^{x}

= \int \frac{9}{e ^{x}} + 6 + 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{9}{e ^{x}} d x + \int 6 d x + \int e^{x} d x

\int \frac{9}{e ^{x}} d x = - \frac{9}{e ^{x}}

\int 6 d x = 6 x

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

= - \frac{9}{e ^{x}} + 6 x + e^{x}

Add a constant to the solution

= - \frac{9}{e ^{x}} + 6 x + e^{x} + C

\frac{\partial}{\partial x} (- y sin (x))

\frac{d}{d x} (\frac{sin ( 2 x )}{tan ( x )})

\frac{1}{( x + 2 )}

\int_{2}^{3} t^{3} d t

What is the integral of ((3+e^x)^2)/(e^x) ?
The integral of ((3+e^x)^2)/(e^x) is -9/(e^x)+6x+e^x+C