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

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

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

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

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

Solution steps

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

Apply Integration By Parts

: \frac{x}{e ^{\frac{x}{2}}} - \int - \frac{1}{2} e^{- \frac{x}{2}} x d x

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

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

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

Add a constant to the solution

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

\int \frac{sin ^{5} ( 2 x )}{cos ^{2} ( 2 x )} d x

\frac{d}{d y} (e^{2 x y})

\int \frac{x ^{3}}{1 - x ^{8}} d x

f (x) = 3 - 4 x

y = 4 cos (x)

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