What is the integral of (9/(sqrt(x))+9sqrt(x)) ?

The integral of (9/(sqrt(x))+9sqrt(x)) is 18sqrt(x)+6x^{3/2}+C

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integral of (9/(sqrt(x))+9sqrt(x))

\int (\frac{9}{x} + 9 x) d x

18 x + 6 x^{\frac{3}{2}} + C

Solution steps

\int \frac{9}{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 \frac{9}{x} d x + \int 9 x d x

\int \frac{9}{x} d x = 18 x

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

= 18 x + 6 x^{\frac{3}{2}}

Add a constant to the solution

= 18 x + 6 x^{\frac{3}{2}} + C

f (x) = 9 x + 13 x e^{x}

\frac{\partial}{\partial z} (x^{2} e^{yz})

\frac{\partial}{\partial x} (sin (z) cos (x y))

x \to 0 lim (\frac{x ^{3} + 5 x}{x})

\int_{0}^{x} \frac{1}{t ^{2} - 25} d t

What is the integral of (9/(sqrt(x))+9sqrt(x)) ?
The integral of (9/(sqrt(x))+9sqrt(x)) is 18sqrt(x)+6x^{3/2}+C