Calculus Examples

$3^{x + 1} d x$

Step 1

Write $3^{x + 1} d x$ as a function.

$f (x) = 3^{x + 1} d x$

Step 2

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

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

Step 3

Set up the integral to solve.

$F (x) = \int 3^{x + 1} d x d x$

Step 4

Since $d$ is constant with respect to $x$ , move $d$ out of the integral.

$d \int 3^{x + 1} x d x$

Step 5

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = x$ and $d v = 3^{x + 1}$ .

$d (x (\frac{1}{\ln (3)} \cdot 3^{x + 1}) - \int \frac{1}{\ln (3)} \cdot 3^{x + 1} d x)$

Step 6

Simplify.

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Step 6.1

Combine $\frac{1}{\ln (3)}$ and $3^{x + 1}$ .

$d (x \frac{3^{x + 1}}{\ln (3)} - \int \frac{1}{\ln (3)} \cdot 3^{x + 1} d x)$

Step 6.2

Combine $x$ and $\frac{3^{x + 1}}{\ln (3)}$ .

$d (\frac{x \cdot 3^{x + 1}}{\ln (3)} - \int \frac{1}{\ln (3)} \cdot 3^{x + 1} d x)$

Step 6.3

Combine $\frac{1}{\ln (3)}$ and $3^{x + 1}$ .

$d (\frac{x \cdot 3^{x + 1}}{\ln (3)} - \int \frac{3^{x + 1}}{\ln (3)} d x)$

Step 7

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

$d (\frac{x \cdot 3^{x + 1}}{\ln (3)} - (\frac{1}{\ln (3)} \int 3^{x + 1} d x))$

Step 8

Let $u = x + 1$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 8.1

Let $u = x + 1$ . Find $\frac{d u}{d x}$ .

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Step 8.1.1

Differentiate $x + 1$ .

$\frac{d}{d x} [x + 1]$

Step 8.1.2

By the Sum Rule, the derivative of $x + 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 8.1.3

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

$1 + \frac{d}{d x} [1]$

Step 8.1.4

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$1 + 0$

Step 8.1.5

Add $1$ and $0$ .

$1$

Step 8.2

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

$d (\frac{x \cdot 3^{x + 1}}{\ln (3)} - \frac{1}{\ln (3)} \int 3^{u} d u)$

Step 9

The integral of $3^{u}$ with respect to $u$ is $\frac{3^{u}}{\ln (3)}$ .

$d (\frac{x \cdot 3^{x + 1}}{\ln (3)} - \frac{1}{\ln (3)} (\frac{3^{u}}{\ln (3)} + C))$

Step 10

Simplify.

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Step 10.1

Rewrite $d (\frac{x \cdot 3^{x + 1}}{\ln (3)} - \frac{1}{\ln (3)} (\frac{3^{u}}{\ln (3)} + C))$ as $d (\frac{x \cdot 3^{x + 1}}{\ln (3)} - \frac{3^{u}}{\ln^{2} (3)}) + C$ .

$d (\frac{x \cdot 3^{x + 1}}{\ln (3)} - \frac{3^{u}}{\ln^{2} (3)}) + C$

Step 10.2

Simplify.

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Step 10.2.1

To write $\frac{x \cdot 3^{x + 1}}{\ln (3)}$ as a fraction with a common denominator, multiply by $\frac{\ln (3)}{\ln (3)}$ .

$d (\frac{x \cdot 3^{x + 1}}{\ln (3)} \cdot \frac{\ln (3)}{\ln (3)} - \frac{3^{u}}{\ln^{2} (3)}) + C$

Step 10.2.2

Write each expression with a common denominator of $\ln^{2} (3)$ , by multiplying each by an appropriate factor of $1$ .

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Step 10.2.2.1

Multiply $\frac{x \cdot 3^{x + 1}}{\ln (3)}$ by $\frac{\ln (3)}{\ln (3)}$ .

$d (\frac{x \cdot 3^{x + 1} \ln (3)}{\ln (3) \ln (3)} - \frac{3^{u}}{\ln^{2} (3)}) + C$

Step 10.2.2.2

Raise $\ln (3)$ to the power of $1$ .

$d (\frac{x \cdot 3^{x + 1} \ln (3)}{\ln^{1} (3) \ln (3)} - \frac{3^{u}}{\ln^{2} (3)}) + C$

Step 10.2.2.3

Raise $\ln (3)$ to the power of $1$ .

$d (\frac{x \cdot 3^{x + 1} \ln (3)}{\ln^{1} (3) \ln^{1} (3)} - \frac{3^{u}}{\ln^{2} (3)}) + C$

Step 10.2.2.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$d (\frac{x \cdot 3^{x + 1} \ln (3)}{{\ln (3)}^{1 + 1}} - \frac{3^{u}}{\ln^{2} (3)}) + C$

Step 10.2.2.5

Add $1$ and $1$ .

$d (\frac{x \cdot 3^{x + 1} \ln (3)}{\ln^{2} (3)} - \frac{3^{u}}{\ln^{2} (3)}) + C$

Step 10.2.3

Combine the numerators over the common denominator.

$d \frac{x \cdot 3^{x + 1} \ln (3) - 3^{u}}{\ln^{2} (3)} + C$

Step 10.2.4

Combine $d$ and $\frac{x \cdot 3^{x + 1} \ln (3) - 3^{u}}{\ln^{2} (3)}$ .

$\frac{d (x \cdot 3^{x + 1} \ln (3) - 3^{u})}{\ln^{2} (3)} + C$

Step 11

Replace all occurrences of $u$ with $x + 1$ .

$\frac{1}{\ln^{2} (3)} d (x \cdot 3^{x + 1} \ln (3) - 3^{x + 1}) + C$

Step 12

The answer is the antiderivative of the function $f (x) = 3^{x + 1} d x$ .

$F (x) =$ $\frac{1}{\ln^{2} (3)} d (x \cdot 3^{x + 1} \ln (3) - 3^{x + 1}) + C$