What is the integral of (-9+8/(sqrt(x))) ?

The integral of (-9+8/(sqrt(x))) is -9x+16sqrt(x)+C

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integral of (-9+8/(sqrt(x)))

\int (- 9 + \frac{8}{x}) d x

- 9 x + 16 x + C

Solution steps

\int - 9 + \frac{8}{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 d x + \int \frac{8}{x} d x

\int 9 d x = 9 x

\int \frac{8}{x} d x = 16 x

= - 9 x + 16 x

Add a constant to the solution

= - 9 x + 16 x + C

\int \frac{18}{x ( x ^{2} - 9 )} d x

\int (18 x + 8) d x

\int \frac{3}{a - b x} d x

\int \frac{x}{x ^{2} x ^{2} - 1} d x

\int + \frac{1}{csc ( 2 x ) - cot ( 2 x )} d x

What is the integral of (-9+8/(sqrt(x))) ?
The integral of (-9+8/(sqrt(x))) is -9x+16sqrt(x)+C