What is the integral of ((e^{x/2}+5))/(4e^{x/2)} ?

The integral of ((e^{x/2}+5))/(4e^{x/2)} is 1/4 (x+5(x/(e^{x/2)}-e^{-x/2}(x+2)))+C

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integral of ((e^{x/2}+5))/(4e^{x/2)}

\int \frac{( e ^{\frac{x}{2}} + 5 )}{4 e ^{\frac{x}{2}}} d x

\frac{1}{4} (x + 5 (\frac{x}{e ^{\frac{x}{2}}} - e^{- \frac{x}{2}} (x + 2))) + C

Solution steps

\int \frac{e ^{\frac{x}{2}} + 5}{4 e ^{\frac{x}{2}}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{e ^{\frac{x}{2}} + 5}{e ^{\frac{x}{2}}} d x

Expand

\frac{e ^{\frac{x}{2}} + 5}{e ^{\frac{x}{2}}} : 1 + \frac{5}{e ^{\frac{x}{2}}}

= \frac{1}{4} \cdot \int 1 + \frac{5}{e ^{\frac{x}{2}}} 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

= \frac{1}{4} (\int 1 d x + \int \frac{5}{e ^{\frac{x}{2}}} d x)

\int 1 d x = x

\int \frac{5}{e ^{\frac{x}{2}}} d x = 5 (\frac{x}{e ^{\frac{x}{2}}} - e^{- \frac{x}{2}} (x + 2))

= \frac{1}{4} (x + 5 (\frac{x}{e ^{\frac{x}{2}}} - e^{- \frac{x}{2}} (x + 2)))

Add a constant to the solution

= \frac{1}{4} (x + 5 (\frac{x}{e ^{\frac{x}{2}}} - e^{- \frac{x}{2}} (x + 2))) + C

\frac{1}{x}, 9

\int_{0}^{2} (x^{2} - 4) d x

\frac{d y}{d x} = \frac{( y - 1 )}{x ^{2}}

6 cos (2 x)

\int_{1}^{t} \frac{21}{x ^{3}} d x

What is the integral of ((e^{x/2}+5))/(4e^{x/2)} ?
The integral of ((e^{x/2}+5))/(4e^{x/2)} is 1/4 (x+5(x/(e^{x/2)}-e^{-x/2}(x+2)))+C