What is the integral of ((4+e^x)^2)/(e^x) ?

The integral of ((4+e^x)^2)/(e^x) is -(16)/(e^x)+8x+e^x+C

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integral of ((4+e^x)^2)/(e^x)

\int \frac{( 4 + e ^{x} ) ^{2}}{e ^{x}} d x

- \frac{16}{e ^{x}} + 8 x + e^{x} + C

Solution steps

\int \frac{( 4 + e ^{x} ) ^{2}}{e ^{x}} d x

Expand

\frac{( 4 + e ^{x} ) ^{2}}{e ^{x}} : \frac{16}{e ^{x}} + 8 + e^{x}

= \int \frac{16}{e ^{x}} + 8 + 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{16}{e ^{x}} d x + \int 8 d x + \int e^{x} d x

\int \frac{16}{e ^{x}} d x = - \frac{16}{e ^{x}}

\int 8 d x = 8 x

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

= - \frac{16}{e ^{x}} + 8 x + e^{x}

Add a constant to the solution

= - \frac{16}{e ^{x}} + 8 x + e^{x} + C

y^{^{''}} + 4 y^{^{'}} + 7 y = 0, y (0) = - 2, y^{^{'}} (0) = 4

x \to 0 - lim (e^{2 x})

\frac{d y}{d x} = y^{2} sin (x)

\frac{d r}{d t} = - r^{3}

1600 y^{^{''}} - 9 y = 0, y (0) = 7, y^{^{'}} (0) = 6

What is the integral of ((4+e^x)^2)/(e^x) ?
The integral of ((4+e^x)^2)/(e^x) is -(16)/(e^x)+8x+e^x+C