What is the integral of (sqrt(x)+9/x-4e^x) ?

The integral of (sqrt(x)+9/x-4e^x) is 2/3 x^{3/2}+9ln|x|-4e^x+C

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integral of (sqrt(x)+9/x-4e^x)

\int (x + \frac{9}{x} - 4 e^{x}) d x

\frac{2}{3} x^{\frac{3}{2}} + 9 ln ∣ x ∣ - 4 e^{x} + C

Solution steps

\int x + \frac{9}{x} - 4 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 x d x + \int \frac{9}{x} d x - \int 4 e^{x} d x

\int x d x = \frac{2}{3} x^{\frac{3}{2}}

\int \frac{9}{x} d x = 9 ln ∣ x ∣

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

= \frac{2}{3} x^{\frac{3}{2}} + 9 ln ∣ x ∣ - 4 e^{x}

Add a constant to the solution

= \frac{2}{3} x^{\frac{3}{2}} + 9 ln ∣ x ∣ - 4 e^{x} + C

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What is the integral of (sqrt(x)+9/x-4e^x) ?
The integral of (sqrt(x)+9/x-4e^x) is 2/3 x^{3/2}+9ln|x|-4e^x+C