What is the integral of 1/(sqrt(6r-7)) ?

The integral of 1/(sqrt(6r-7)) is 1/3 sqrt(6r-7)+C

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integral of 1/(sqrt(6r-7))

\int \frac{1}{6 r - 7} d r

\frac{1}{3} 6 r - 7 + C

Solution steps

\int \frac{1}{6 r - 7} d r

Apply u-substitution

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

= \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

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

Apply the Power Rule

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

= \frac{1}{6} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 6 r - 7

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

Simplify

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

= \frac{1}{3} 6 r - 7

Add a constant to the solution

= \frac{1}{3} 6 r - 7 + C

\int sec^{2} (x) csc^{4} (x) d x

y = 2 x - \frac{1}{2}

\int 5 x^{- 2} d x

\int 5 x^{6} d x

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

What is the integral of 1/(sqrt(6r-7)) ?
The integral of 1/(sqrt(6r-7)) is 1/3 sqrt(6r-7)+C