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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u ^{\frac{1}{2}} ( u + 1 )}{4} d u

= \int \frac{u ^{\frac{1}{2}} ( 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^{\frac{1}{2}} (u + 1) d u

Expand

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

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

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

\int u^{\frac{1}{2}} 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 = 2 x - 1

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

Simplify

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

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