What is the integral of 7sqrt(4x+8) ?

The integral of 7sqrt(4x+8) is 28/3 (x+2)^{3/2}+C

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integral of 7sqrt(4x+8)

\int 7 4 x + 8 d x

\frac{28}{3} (x + 2)^{\frac{3}{2}} + C

Solution steps

\int 7 4 x + 8 d x

Take the constant out:

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

= 7 \cdot \int 4 x + 8 d x

Apply u-substitution

= 7 \cdot \int \frac{u}{4} d u

Take the constant out:

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

= 7 \cdot \frac{1}{4} \cdot \int u d u

Apply the Power Rule

= 7 \cdot \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 4 x + 8

= 7 \cdot \frac{1}{4} \cdot \frac{2}{3} (4 x + 8)^{\frac{3}{2}}

Simplify

7 \cdot \frac{1}{4} \cdot \frac{2}{3} (4 x + 8)^{\frac{3}{2}} : \frac{28}{3} (x + 2)^{\frac{3}{2}}

= \frac{28}{3} (x + 2)^{\frac{3}{2}}

Add a constant to the solution

= \frac{28}{3} (x + 2)^{\frac{3}{2}} + C

What is the integral of 7sqrt(4x+8) ?
The integral of 7sqrt(4x+8) is 28/3 (x+2)^{3/2}+C