Find the Function f(x)=(3x+2)^4

$f (x) = {(3 x + 2)}^{4}$

The function $F (x)$ can be found by evaluating the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Let $u = 3 x + 2$ . Then $d u = 3 d x$ , so $\frac{1}{3} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

Let $u = 3 x + 2$ . Find $\frac{d u}{d x}$ .

Differentiate $3 x + 2$ .

$\frac{d}{d x} [3 x + 2]$

By the Sum Rule, the derivative of $3 x + 2$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [2]$ .

$\frac{d}{d x} [3 x] + \frac{d}{d x} [2]$

Evaluate $\frac{d}{d x} [3 x]$ .

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$3 \frac{d}{d x} [x] + \frac{d}{d x} [2]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$3 \cdot 1 + \frac{d}{d x} [2]$

Multiply $3$ by $1$ .

$3 + \frac{d}{d x} [2]$

Differentiate using the Constant Rule.

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$3 + 0$

Add $3$ and $0$ .

$3$

Rewrite the problem using $u$ and $d u$ .

$\int u^{4} \frac{1}{3} d u$

Combine $u^{4}$ and $\frac{1}{3}$ .

$\int \frac{u^{4}}{3} d u$

Since $\frac{1}{3}$ is constant with respect to $u$ , move $\frac{1}{3}$ out of the integral.

$\frac{1}{3} \int u^{4} d u$

By the Power Rule, the integral of $u^{4}$ with respect to $u$ is $\frac{1}{5} u^{5}$ .

$\frac{1}{3} (\frac{1}{5} u^{5} + C)$

Simplify.

Rewrite $\frac{1}{3} (\frac{1}{5} u^{5} + C)$ as $\frac{1}{3} \cdot \frac{1}{5} u^{5} + C$ .

$\frac{1}{3} \cdot \frac{1}{5} u^{5} + C$

Simplify.

Multiply $\frac{1}{3}$ by $\frac{1}{5}$ .

$\frac{1}{3 \cdot 5} u^{5} + C$

Multiply $3$ by $5$ .

$\frac{1}{15} u^{5} + C$

Replace all occurrences of $u$ with $3 x + 2$ .

$\frac{1}{15} {(3 x + 2)}^{5} + C$

The function $F$ if derived from the integral of the derivative of the function. This is valid by the fundamental theorem of calculus.

$F (x) = \frac{1}{15} \cdot {(3 x + 2)}^{5} + C$