What is the integral of sqrt(6x+4) ?

The integral of sqrt(6x+4) is 1/9 (6x+4)^{3/2}+C

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integral of sqrt(6x+4)

\int 6 x + 4 d x

\frac{1}{9} (6 x + 4)^{\frac{3}{2}} + C

Solution steps

\int 6 x + 4 d x

Apply u-substitution

: \int \frac{u}{6} d u

= \int \frac{u}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{6} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 6 x + 4

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

Simplify

\frac{1}{6} \cdot \frac{2}{3} (6 x + 4)^{\frac{3}{2}} : \frac{1}{9} (6 x + 4)^{\frac{3}{2}}

= \frac{1}{9} (6 x + 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{9} (6 x + 4)^{\frac{3}{2}} + C

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

\frac{\partial}{\partial y} (1 - \frac{3}{y} + x)

f (x) = (x + 1)^{x + 1}

y = sin (9 x) + cos (7 x)

What is the integral of sqrt(6x+4) ?
The integral of sqrt(6x+4) is 1/9 (6x+4)^{3/2}+C