What is the integral of 7/(sqrt(5-6x)) ?

The integral of 7/(sqrt(5-6x)) is -7/3 sqrt(5-6x)+C

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integral of 7/(sqrt(5-6x))

\int \frac{7}{5 - 6 x} d x

- \frac{7}{3} 5 - 6 x + C

Solution steps

\int \frac{7}{5 - 6 x} d x

Take the constant out:

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

= 7 \cdot \int \frac{1}{5 - 6 x} d x

Apply u-substitution

: \int - \frac{1}{6 u} d u

= 7 \cdot \int - \frac{1}{6 u} d u

Take the constant out:

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

= 7 (- \frac{1}{6} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

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

= 7 (- \frac{1}{6} \cdot 2 u^{\frac{1}{2}})

Substitute back

u = 5 - 6 x

= 7 (- \frac{1}{6} \cdot 2 (5 - 6 x)^{\frac{1}{2}})

Simplify

7 (- \frac{1}{6} \cdot 2 (5 - 6 x)^{\frac{1}{2}}) : - \frac{7}{3} 5 - 6 x

= - \frac{7}{3} 5 - 6 x

Add a constant to the solution

= - \frac{7}{3} 5 - 6 x + C

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\frac{d}{d x} (ln (e^{x} - e^{- x}))

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

\int \frac{y}{16 + y ^{2}} d y

x^{5} + y^{4} x^{6} + 4 y^{^{'}} = 0

What is the integral of 7/(sqrt(5-6x)) ?
The integral of 7/(sqrt(5-6x)) is -7/3 sqrt(5-6x)+C