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

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

Popular >

integral of sqrt(4x-1)

\int 4 x - 1 d x

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

Solution steps

\int 4 x - 1 d x

Apply u-substitution

: \int \frac{u}{4} d u

= \int \frac{u}{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 d u

Apply the Power Rule

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

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

Substitute back

u = 4 x - 1

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

Simplify

\frac{1}{4} \cdot \frac{2}{3} (4 x - 1)^{\frac{3}{2}} : \frac{1}{6} (4 x - 1)^{\frac{3}{2}}

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

Add a constant to the solution

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

\frac{d}{d x} ((x^{2} + 1)^{s i n (x)})

\int_{0}^{1} 2 (1 + x)^{5} d x

0.8 e^{x} arctan (x)

\int \frac{( x + 1 )}{x - 1} d x

\frac{d ^{3}}{d x ^{3}} (\frac{8}{x ^{2}})

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