What is the integral of (1-3x)^{1/2} ?

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

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integral of (1-3x)^{1/2}

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{u ^{\frac{1}{2}}}{3} d u

= \int - \frac{u ^{\frac{1}{2}}}{3} d u

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 1 - 3 x

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

Simplify

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

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

Add a constant to the solution

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

x^{2}

\frac{2 s + 1}{s ^{2}}

x y^{^{'}} + y = x y^{3}

n = 0 \sum \infty (\frac{3}{8})^{\frac{n}{2}}

\frac{d v}{d x} = x

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