What is the integral of (2x+8)^3 ?

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

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integral of (2x+8)^3

integral

(2 x + 8)^{3}

\frac{1}{8} (2 x + 8)^{4} + C

Solution steps

\int (2 x + 8)^{3} d x

Apply u-substitution

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

= \int \frac{u ^{3}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

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

Substitute back

u = 2 x + 8

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

Simplify

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

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

Add a constant to the solution

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

x^{2} arcsin (x)

\frac{1}{1 + x ^{3}}

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

\frac{1}{( x ^{2} + 1 ) ^{\frac{1}{2}}}

ln (x - 2)