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

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

y = 2 sin (π x - y), (1, 0)

y^{^{''}} - 5 y^{^{'}} + 6 y = e^{6 x}

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

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