What is the integral of 1/(sqrt(9s-8)) ?

The integral of 1/(sqrt(9s-8)) is 2/9 sqrt(9s-8)+C

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integral of 1/(sqrt(9s-8))

\int \frac{1}{9 s - 8} d s

\frac{2}{9} 9 s - 8 + C

Solution steps

\int \frac{1}{9 s - 8} d s

Apply u-substitution

: \int \frac{1}{9 u} d u

= \int \frac{1}{9 u} d u

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 9 s - 8

= \frac{1}{9} \cdot 2 (9 s - 8)^{\frac{1}{2}}

Simplify

\frac{1}{9} \cdot 2 (9 s - 8)^{\frac{1}{2}} : \frac{2}{9} 9 s - 8

= \frac{2}{9} 9 s - 8

Add a constant to the solution

= \frac{2}{9} 9 s - 8 + C

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What is the integral of 1/(sqrt(9s-8)) ?
The integral of 1/(sqrt(9s-8)) is 2/9 sqrt(9s-8)+C