What is the integral of sqrt(3u+4) ?

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

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integral of sqrt(3u+4)

\int 3 u + 4 d u

\frac{2}{9} (3 u + 4)^{\frac{3}{2}} + C

Solution steps

\int 3 u + 4 d u

Apply u-substitution

= \int \frac{v}{3} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int v d v

Apply the Power Rule

= \frac{1}{3} \cdot \frac{2}{3} v^{\frac{3}{2}}

Substitute back

v = 3 u + 4

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

Simplify

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

= \frac{2}{9} (3 u + 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{9} (3 u + 4)^{\frac{3}{2}} + C

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