Calculus Examples

$f (x) = \frac{1}{\sqrt{7 x - 9}}$

Step 1

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 2

Set up the integral to solve.

$F (x) = \int \frac{1}{\sqrt{7 x - 9}} d x$

Step 3

Let $u = 7 x - 9$ . Then $d u = 7 d x$ , so $\frac{1}{7} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 3.1

Let $u = 7 x - 9$ . Find $\frac{d u}{d x}$ .

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Step 3.1.1

Differentiate $7 x - 9$ .

$\frac{d}{d x} [7 x - 9]$

Step 3.1.2

By the Sum Rule, the derivative of $7 x - 9$ with respect to $x$ is $\frac{d}{d x} [7 x] + \frac{d}{d x} [- 9]$ .

$\frac{d}{d x} [7 x] + \frac{d}{d x} [- 9]$

Step 3.1.3

Evaluate $\frac{d}{d x} [7 x]$ .

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Step 3.1.3.1

Since $7$ is constant with respect to $x$ , the derivative of $7 x$ with respect to $x$ is $7 \frac{d}{d x} [x]$ .

$7 \frac{d}{d x} [x] + \frac{d}{d x} [- 9]$

Step 3.1.3.2

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

$7 \cdot 1 + \frac{d}{d x} [- 9]$

Step 3.1.3.3

Multiply $7$ by $1$ .

$7 + \frac{d}{d x} [- 9]$

Step 3.1.4

Differentiate using the Constant Rule.

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Step 3.1.4.1

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9$ with respect to $x$ is $0$ .

$7 + 0$

Step 3.1.4.2

Add $7$ and $0$ .

$7$

Step 3.2

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

$\int \frac{1}{\sqrt{u}} \cdot \frac{1}{7} d u$

Step 4

Simplify.

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Step 4.1

Multiply $\frac{1}{\sqrt{u}}$ by $\frac{1}{7}$ .

$\int \frac{1}{\sqrt{u} \cdot 7} d u$

Step 4.2

Move $7$ to the left of $\sqrt{u}$ .

$\int \frac{1}{7 \sqrt{u}} d u$

Step 5

Since $\frac{1}{7}$ is constant with respect to $u$ , move $\frac{1}{7}$ out of the integral.

$\frac{1}{7} \int \frac{1}{\sqrt{u}} d u$

Step 6

Apply basic rules of exponents.

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Step 6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{u}$ as $u^{\frac{1}{2}}$ .

$\frac{1}{7} \int \frac{1}{u^{\frac{1}{2}}} d u$

Step 6.2

Move $u^{\frac{1}{2}}$ out of the denominator by raising it to the $- 1$ power.

$\frac{1}{7} \int {(u^{\frac{1}{2}})}^{- 1} d u$

Step 6.3

Multiply the exponents in ${(u^{\frac{1}{2}})}^{- 1}$ .

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Step 6.3.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{1}{7} \int u^{\frac{1}{2} \cdot - 1} d u$

Step 6.3.2

Combine $\frac{1}{2}$ and $- 1$ .

$\frac{1}{7} \int u^{\frac{- 1}{2}} d u$

Step 6.3.3

Move the negative in front of the fraction.

$\frac{1}{7} \int u^{- \frac{1}{2}} d u$

Step 7

By the Power Rule, the integral of $u^{- \frac{1}{2}}$ with respect to $u$ is $2 u^{\frac{1}{2}}$ .

$\frac{1}{7} (2 u^{\frac{1}{2}} + C)$

Step 8

Simplify.

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Step 8.1

Rewrite $\frac{1}{7} (2 u^{\frac{1}{2}} + C)$ as $\frac{1}{7} \cdot 2 u^{\frac{1}{2}} + C$ .

$\frac{1}{7} \cdot 2 u^{\frac{1}{2}} + C$

Step 8.2

Combine $\frac{1}{7}$ and $2$ .

$\frac{2}{7} u^{\frac{1}{2}} + C$

Step 9

Replace all occurrences of $u$ with $7 x - 9$ .

$\frac{2}{7} {(7 x - 9)}^{\frac{1}{2}} + C$

Step 10

The answer is the antiderivative of the function $f (x) = \frac{1}{\sqrt{7 x - 9}}$ .

$F (x) =$ $\frac{2}{7} {(7 x - 9)}^{\frac{1}{2}} + C$