What is the integral of (2x+4)^3 ?

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

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integral of (2x+4)^3

\int (2 x + 4)^{3} d x

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

Solution steps

\int (2 x + 4)^{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 + 4

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (x^{6 y})

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 1, y^{^{'}} (0) = k

(1 + \frac{3}{y} + x) (\frac{d y}{d x}) + y = \frac{3}{x} - 1

(9, - 10), (- 10, 3)

\frac{\partial}{\partial x} (x^{5} + 4^{y} + x^{y})