What is the integral of 6/(sqrt(8-3x)) ?

The integral of 6/(sqrt(8-3x)) is -4sqrt(8-3x)+C

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integral of 6/(sqrt(8-3x))

\int \frac{6}{8 - 3 x} d x

- 4 8 - 3 x + C

Solution steps

\int \frac{6}{8 - 3 x} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{8 - 3 x} d x

Apply u-substitution

= 6 \cdot \int - \frac{1}{3 u} d u

Take the constant out:

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

= 6 (- \frac{1}{3} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

= 6 (- \frac{1}{3} \cdot 2 u^{\frac{1}{2}})

Substitute back

u = 8 - 3 x

= 6 (- \frac{1}{3} \cdot 2 (8 - 3 x)^{\frac{1}{2}})

Simplify

6 (- \frac{1}{3} \cdot 2 (8 - 3 x)^{\frac{1}{2}}) : - 4 8 - 3 x

= - 4 8 - 3 x

Add a constant to the solution

= - 4 8 - 3 x + C

What is the integral of 6/(sqrt(8-3x)) ?
The integral of 6/(sqrt(8-3x)) is -4sqrt(8-3x)+C