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

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

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integral of 2xsqrt(2x-1)

\int 2 x 2 x - 1 d x

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

Solution steps

\int 2 x 2 x - 1 d x

Take the constant out:

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

= 2 \cdot \int x 2 x - 1 d x

Apply u-substitution

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

= 2 \cdot \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

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

Expand

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

= 2 \cdot \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

= 2 \cdot \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}}

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

Substitute back

u = 2 x - 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{1}{6 x ^{\frac{5}{6}}})

(1, 1), (4, - \frac{1}{2})

\frac{\partial}{\partial x} (r)

y = \frac{- 8 x}{x ^{2} + 1}, (- 1, 4)

x^{2} (240 - \frac{x}{3})

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