What is the integral of (e^{4x}-3/(e^{x/2)}) ?

The integral of (e^{4x}-3/(e^{x/2)}) is 1/4 e^{4x}-3(x/(e^{x/2)}-e^{-x/2}(x+2))+C

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integral of (e^{4x}-3/(e^{x/2)})

\int (e^{4 x} - \frac{3}{e ^{\frac{x}{2}}}) d x

\frac{1}{4} e^{4 x} - 3 (\frac{x}{e ^{\frac{x}{2}}} - e^{- \frac{x}{2}} (x + 2)) + C

Solution steps

\int e^{4 x} - \frac{3}{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

= \int e^{4 x} d x - \int \frac{3}{e ^{\frac{x}{2}}} d x

\int e^{4 x} d x = \frac{1}{4} e^{4 x}

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

= \frac{1}{4} e^{4 x} - 3 (\frac{x}{e ^{\frac{x}{2}}} - e^{- \frac{x}{2}} (x + 2))

Add a constant to the solution

= \frac{1}{4} e^{4 x} - 3 (\frac{x}{e ^{\frac{x}{2}}} - e^{- \frac{x}{2}} (x + 2)) + C

\frac{d}{d x} ((\frac{x}{2} + 1)^{3})

\int e^{6 x} sin (x) d x

f (x) = 1.5 x^{7} - \frac{x ^{4}}{3} + π x - e x^{2} + 2

f (x) = x^{3} - 4 x^{2}

\int_{0}^{\frac{π}{4}} sin (x) d x

What is the integral of (e^{4x}-3/(e^{x/2)}) ?
The integral of (e^{4x}-3/(e^{x/2)}) is 1/4 e^{4x}-3(x/(e^{x/2)}-e^{-x/2}(x+2))+C