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

The integral of (3x+4)^3 is 1/12 (3x+4)^4+C

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integral of (3x+4)^3

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

\frac{1}{12} (3 x + 4)^{4} + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 3 x + 4

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

Simplify

\frac{1}{3} \cdot \frac{( 3 x + 4 ) ^{4}}{4} : \frac{1}{12} (3 x + 4)^{4}

= \frac{1}{12} (3 x + 4)^{4}

Add a constant to the solution

= \frac{1}{12} (3 x + 4)^{4} + C

n = 0 \sum \infty 16 + 24 + 36 + 54

\frac{\partial}{\partial x} (e^{5 x y} 5 y)

\frac{1}{x ^{2} - 1}

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

\frac{10}{s ^{2}} - 1