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

: \int - \frac{1}{3 u} d u

= 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

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

= 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

f (x) = \frac{4}{x}, at x = - 9

x \to \infty lim (ln (x!))

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

\frac{d}{d x} ((3 x^{2} - 6 x + 6) e^{x})

n = 1 \sum \infty \frac{3}{4 ^{n}}

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