What is the integral of sqrt(1-6x) ?

The integral of sqrt(1-6x) is -1/9 (1-6x)^{3/2}+C

Popular >

integral of sqrt(1-6x)

\int 1 - 6 x d x

- \frac{1}{9} (1 - 6 x)^{\frac{3}{2}} + C

Solution steps

\int 1 - 6 x d x

Apply u-substitution

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

= \int - \frac{u}{6} 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 u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

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

Substitute back

u = 1 - 6 x

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

Simplify

- \frac{1}{6} \cdot \frac{2}{3} (1 - 6 x)^{\frac{3}{2}} : - \frac{1}{9} (1 - 6 x)^{\frac{3}{2}}

= - \frac{1}{9} (1 - 6 x)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{9} (1 - 6 x)^{\frac{3}{2}} + C

\int \frac{e ^{a r c s i n (x)}}{1 - x ^{2}} d x

\frac{d}{d x} (\frac{12}{x + 2} - 3)

\frac{d}{d x} (\frac{x ^{2} - 4 x}{x + 1})

\int - \frac{x ^{2} + 11 x + 28}{( x + 2 ) ^{2}} d x

\frac{d}{d x} (\frac{1}{x ^{2} + 5})

What is the integral of sqrt(1-6x) ?
The integral of sqrt(1-6x) is -1/9 (1-6x)^{3/2}+C