What is the integral of 1/(sqrt(9x-7)) ?

The integral of 1/(sqrt(9x-7)) is 2/9 sqrt(9x-7)+C

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integral of 1/(sqrt(9x-7))

\int \frac{1}{9 x - 7} d x

\frac{2}{9} 9 x - 7 + C

Solution steps

\int \frac{1}{9 x - 7} d x

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 x - 7

= \frac{1}{9} \cdot 2 (9 x - 7)^{\frac{1}{2}}

Simplify

\frac{1}{9} \cdot 2 (9 x - 7)^{\frac{1}{2}} : \frac{2}{9} 9 x - 7

= \frac{2}{9} 9 x - 7

Add a constant to the solution

= \frac{2}{9} 9 x - 7 + C

y = y^{^{'}}

\frac{d}{d y} (e^{- \frac{1}{y ^{3}}})

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

11 x

y^{^{''}} - y^{^{'}} - 12 y = e^{4 x}

What is the integral of 1/(sqrt(9x-7)) ?
The integral of 1/(sqrt(9x-7)) is 2/9 sqrt(9x-7)+C