What is the integral of sqrt(6x+3) ?

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

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integral of sqrt(6x+3)

\int 6 x + 3 d x

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

Solution steps

\int 6 x + 3 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 + 3

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

Simplify

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

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{1}{e ^{n^{3}}}

\int \frac{1}{( 2 x - 1 ) ^{3}} d x

x \to - 6 lim (\frac{x ^{2} + 6}{x - 6})

\frac{d}{d x} (e^{2 - 3 x})

\frac{d}{d x} (3 cosh (x))

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