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

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

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integral of sqrt(1-2x)

\int 1 - 2 x d x

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

Solution steps

\int 1 - 2 x d x

Apply u-substitution

: \int - \frac{u}{2} d u

= \int - \frac{u}{2} d u

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 1 - 2 x

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

Simplify

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

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

Add a constant to the solution

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

\int sin (x) + sinh (x) d x

\frac{d y}{d x} = \frac{( 10 + 15 x )}{x y ^{2}}, y (1) = 3

\frac{\partial}{\partial x} (x \cdot y)

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

\frac{\partial}{\partial x} (16 - x^{2} - 4 y)

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