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

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

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integral of 6*sqrt(x+4)

\int 6 \cdot x + 4 d x

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

Solution steps

\int 6 x + 4 d x

Take the constant out:

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

= 6 \cdot \int x + 4 d x

Apply u-substitution

: \int u d u

= 6 \cdot \int u d u

Apply the Power Rule

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

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

Substitute back

u = x + 4

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (sin (x + 2 y + 3 z))

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

x \to 4 lim (\frac{7}{( x - 4 )})

\frac{\partial}{\partial x} (y ln (x y))

x \to 2 lim (\frac{6}{( x - 2 ) ^{3}})

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