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

The integral of (9e^x-9sqrt(x)) is 9e^x-6x^{3/2}+C

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integral of (9e^x-9sqrt(x))

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

9 e^{x} - 6 x^{\frac{3}{2}} + C

Solution steps

\int 9 e^{x} - 9 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 9 e^{x} d x - \int 9 x d x

\int 9 e^{x} d x = 9 e^{x}

\int 9 x d x = 6 x^{\frac{3}{2}}

= 9 e^{x} - 6 x^{\frac{3}{2}}

Add a constant to the solution

= 9 e^{x} - 6 x^{\frac{3}{2}} + C

x^{3} e^{6 x^{3} + 8 x^{2} - 7}

\frac{d}{d x} (arcsin (x^{3}))

f (x) = \frac{f ( x )}{x ^{2} + 2}

\int_{1}^{16} \frac{4}{x} d x

x \to 0 lim (\frac{2 x}{x ^{2} - 5 x})

What is the integral of (9e^x-9sqrt(x)) ?
The integral of (9e^x-9sqrt(x)) is 9e^x-6x^{3/2}+C