What is the integral of 1/(sqrt(2-8x)) ?

The integral of 1/(sqrt(2-8x)) is -1/4 sqrt(2-8x)+C

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integral of 1/(sqrt(2-8x))

\int \frac{1}{2 - 8 x} d x

- \frac{1}{4} 2 - 8 x + C

Solution steps

\int \frac{1}{2 - 8 x} d x

Apply u-substitution

: \int - \frac{1}{8 u} d u

= \int - \frac{1}{8 u} d u

Take the constant out:

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

= - \frac{1}{8} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= - \frac{1}{8} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 2 - 8 x

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

Simplify

- \frac{1}{8} \cdot 2 (2 - 8 x)^{\frac{1}{2}} : - \frac{1}{4} 2 - 8 x

= - \frac{1}{4} 2 - 8 x

Add a constant to the solution

= - \frac{1}{4} 2 - 8 x + C

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What is the integral of 1/(sqrt(2-8x)) ?
The integral of 1/(sqrt(2-8x)) is -1/4 sqrt(2-8x)+C