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

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

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

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

- 9 p + 16 p + C

Solution steps

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

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 p + \int \frac{8}{p} d p

\int 9 d p = 9 p

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

= - 9 p + 16 p

Add a constant to the solution

= - 9 p + 16 p + C

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