What is the integral of xsqrt(4x-1) ?

The integral of xsqrt(4x-1) is 1/60 (6x+1)(4x-1)^{3/2}+C

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integral of xsqrt(4x-1)

\int x 4 x - 1 d x

\frac{1}{60} (6 x + 1) (4 x - 1)^{\frac{3}{2}} + C

Solution steps

\int x 4 x - 1 d x

Apply u-substitution

= \int \frac{u ( u + 1 )}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int u (u + 1) d u

Expand

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

= \frac{1}{16} \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}{16} (\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}{16} (\frac{2}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 4 x - 1

= \frac{1}{16} (\frac{2}{5} (4 x - 1)^{\frac{5}{2}} + \frac{2}{3} (4 x - 1)^{\frac{3}{2}})

Simplify

\frac{1}{16} (\frac{2}{5} (4 x - 1)^{\frac{5}{2}} + \frac{2}{3} (4 x - 1)^{\frac{3}{2}}) : \frac{1}{60} (6 x + 1) (4 x - 1)^{\frac{3}{2}}

= \frac{1}{60} (6 x + 1) (4 x - 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{60} (6 x + 1) (4 x - 1)^{\frac{3}{2}} + C

d er i v a t i v e \frac{e ^{x}}{x ^{2} + 14}

\int_{1}^{2} \frac{7}{x ^{3} + 4 x} d x

\frac{\partial}{\partial y} (sin (y) - y sin (x))

\frac{d}{d x} (e^{x x^{2} + 1})

y^{^{''}} - 2 y + 4 y = 0

What is the integral of xsqrt(4x-1) ?
The integral of xsqrt(4x-1) is 1/60 (6x+1)(4x-1)^{3/2}+C