What is the integral of 9/(sqrt(t)) ?

The integral of 9/(sqrt(t)) is 18sqrt(t)+C

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integral of 9/(sqrt(t))

\int \frac{9}{t} d t

18 t + C

Solution steps

\int \frac{9}{t} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int \frac{1}{t} d t

Apply the Power Rule

: 2 t^{\frac{1}{2}}

= 9 \cdot 2 t^{\frac{1}{2}}

Simplify

9 \cdot 2 t^{\frac{1}{2}} : 18 t

= 18 t

Add a constant to the solution

= 18 t + C

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What is the integral of 9/(sqrt(t)) ?
The integral of 9/(sqrt(t)) is 18sqrt(t)+C