What is the integral of 8sqrt(x/4+1) ?

The integral of 8sqrt(x/4+1) is 8/3 (x+4)^{3/2}+C

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integral of 8sqrt(x/4+1)

\int 8 \frac{x}{4} + 1 d x

\frac{8}{3} (x + 4)^{\frac{3}{2}} + C

Solution steps

\int 8 \frac{x}{4} + 1 d x

Take the constant out:

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

= 8 \cdot \int \frac{x}{4} + 1 d x

Apply u-substitution

: \int u \cdot 4 d u

= 8 \cdot \int u \cdot 4 d u

Take the constant out:

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

= 8 \cdot 4 \cdot \int u d u

Apply the Power Rule

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

= 8 \cdot 4 \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = \frac{x}{4} + 1

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

Simplify

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

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

Add a constant to the solution

= \frac{8}{3} (x + 4)^{\frac{3}{2}} + C

\frac{d}{d x} (8 x^{2} + x)

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

What is the integral of 8sqrt(x/4+1) ?
The integral of 8sqrt(x/4+1) is 8/3 (x+4)^{3/2}+C