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

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

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integral of 2sqrt(2x-1)

\int 2 2 x - 1 d x

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

Solution steps

\int 2 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 2 x - 1 d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 2 x - 1

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{5}{( 3 x - 1 ) ^{5}} d x

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\frac{\partial}{\partial x} (4 x^{2} y^{3} - 4)

y^{^{'}} = 0.2 ln (\frac{3000}{y}) y, y (0) = 1000

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

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