What is the integral of sqrt(6x+2) ?

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

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integral of sqrt(6x+2)

\int 6 x + 2 d x

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

Solution steps

\int 6 x + 2 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 + 2

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

Simplify

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

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

Add a constant to the solution

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

- 3 - 3 x^{2}

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

\frac{d y}{d x} = x (1 + y)

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

\int \frac{3}{x ^{2}} + 4 x d x

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