What is the integral of sqrt(4x-3) ?

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

integral of sqrt(4x-3)

\int 4 x - 3 d x

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

Solution steps

\int 4 x - 3 d x

Apply u-substitution

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

Substitute back

u = 4 x - 3

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{x + 5}{( x + 3 ) ( x - 2 ) ^{2}} d x

\frac{d y}{d t} - 2 t y = - 12 t^{2} e^{t^{2}}, y (0) = - 1

\int (ln (4 x))^{2} d x

y y^{^{'}} + x = 0

\int \frac{r + 3}{r + 3} d r

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