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

The integral of xsqrt(1-2x) is -1/15 (3x+1)(-2x+1)^{3/2}+C

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integral of xsqrt(1-2x)

\int x 1 - 2 x d x

- \frac{1}{15} (3 x + 1) (- 2 x + 1)^{\frac{3}{2}} + C

Solution steps

\int x 1 - 2 x d x

Apply u-substitution

: \int \frac{u ( u - 1 )}{4} d u

= \int \frac{u ( u - 1 )}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u (u - 1) d u

Expand

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

= \frac{1}{4} \cdot \int u^{\frac{3}{2}} - u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} (\int u^{\frac{3}{2}} d u - \int u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

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

= \frac{1}{4} (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 1 - 2 x

= \frac{1}{4} (\frac{2}{5} (1 - 2 x)^{\frac{5}{2}} - \frac{2}{3} (1 - 2 x)^{\frac{3}{2}})

Simplify

\frac{1}{4} (\frac{2}{5} (1 - 2 x)^{\frac{5}{2}} - \frac{2}{3} (1 - 2 x)^{\frac{3}{2}}) : - \frac{1}{15} (3 x + 1) (- 2 x + 1)^{\frac{3}{2}}

= - \frac{1}{15} (3 x + 1) (- 2 x + 1)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{15} (3 x + 1) (- 2 x + 1)^{\frac{3}{2}} + C

\frac{\partial}{\partial x} (\frac{x}{e ^{\frac{x}{y}}})

\frac{d}{d x} (- \frac{2}{( 6 x - 7 x ^{4} ) ^{3}})

\frac{d y}{d x} = lo g_{10} (x)

\frac{d y}{d x} = \frac{x ^{9} - 8 y}{x}

\frac{d}{d x} (∣ 4 x - 20 ∣)

What is the integral of xsqrt(1-2x) ?
The integral of xsqrt(1-2x) is -1/15 (3x+1)(-2x+1)^{3/2}+C