Calculus Examples

$\frac{d f}{d x} = \frac{4 x + x f^{2}}{f - x^{2} f}$

Step 1

Separate the variables.

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Step 1.1

Factor $x$ out of $4 x + x f^{2}$ .

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Step 1.1.1

Factor $x$ out of $4 x$ .

$\frac{d f}{d x} = \frac{x \cdot 4 + x f^{2}}{f - x^{2} f}$

Step 1.1.2

Factor $x$ out of $x f^{2}$ .

$\frac{d f}{d x} = \frac{x \cdot 4 + x (f^{2})}{f - x^{2} f}$

Step 1.1.3

Factor $x$ out of $x \cdot 4 + x (f^{2})$ .

$\frac{d f}{d x} = \frac{x \cdot (4 + f^{2})}{f - x^{2} f}$

Step 1.1.4

Multiply $x$ by $4 + f^{2}$ .

$\frac{d f}{d x} = \frac{x (4 + f^{2})}{f - x^{2} f}$

Step 1.2

Factor.

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Step 1.2.1

Factor $f$ out of $f - x^{2} f$ .

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Step 1.2.1.1

Raise $f$ to the power of $1$ .

$\frac{d f}{d x} = \frac{x (4 + f^{2})}{f^{1} - x^{2} f}$

Step 1.2.1.2

Factor $f$ out of $f^{1}$ .

$\frac{d f}{d x} = \frac{x (4 + f^{2})}{f \cdot 1 - x^{2} f}$

Step 1.2.1.3

Factor $f$ out of $- x^{2} f$ .

$\frac{d f}{d x} = \frac{x (4 + f^{2})}{f \cdot 1 + f (- x^{2})}$

Step 1.2.1.4

Factor $f$ out of $f \cdot 1 + f (- x^{2})$ .

$\frac{d f}{d x} = \frac{x (4 + f^{2})}{f \cdot (1 - x^{2})}$

Step 1.2.1.5

Multiply $f$ by $1 - x^{2}$ .

$\frac{d f}{d x} = \frac{x (4 + f^{2})}{f (1 - x^{2})}$

Step 1.2.2

Rewrite $1$ as $1^{2}$ .

$\frac{d f}{d x} = \frac{x (4 + f^{2})}{f (1^{2} - x^{2})}$

Step 1.2.3

Factor.

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Step 1.2.3.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = x$ .

$\frac{d f}{d x} = \frac{x (4 + f^{2})}{f ((1 + x) (1 - x))}$

Step 1.2.3.2

Remove unnecessary parentheses.

$\frac{d f}{d x} = \frac{x (4 + f^{2})}{f (1 + x) (1 - x)}$

Step 1.3

Regroup factors.

$\frac{d f}{d x} = \frac{x}{(1 + x) (1 - x)} \cdot \frac{4 + f^{2}}{f}$

Step 1.4

Multiply both sides by $\frac{f}{4 + f^{2}}$ .

$\frac{f}{4 + f^{2}} \frac{d f}{d x} = \frac{f}{4 + f^{2}} (\frac{x}{(1 + x) (1 - x)} \cdot \frac{4 + f^{2}}{f})$

Step 1.5

Simplify.

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Step 1.5.1

Multiply $\frac{x}{(1 + x) (1 - x)}$ by $\frac{4 + f^{2}}{f}$ .

$\frac{f}{4 + f^{2}} \frac{d f}{d x} = \frac{f}{4 + f^{2}} \cdot \frac{x (4 + f^{2})}{(1 + x) (1 - x) f}$

Step 1.5.2

Cancel the common factor of $f$ .

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Step 1.5.2.1

Factor $f$ out of $(1 + x) (1 - x) f$ .

$\frac{f}{4 + f^{2}} \frac{d f}{d x} = \frac{f}{4 + f^{2}} \cdot \frac{x (4 + f^{2})}{f ((1 + x) (1 - x))}$

Step 1.5.2.2

Cancel the common factor.

$\frac{f}{4 + f^{2}} \frac{d f}{d x} = \frac{f}{4 + f^{2}} \cdot \frac{x (4 + f^{2})}{f ((1 + x) (1 - x))}$

Step 1.5.2.3

Rewrite the expression.

$\frac{f}{4 + f^{2}} \frac{d f}{d x} = \frac{1}{4 + f^{2}} \cdot \frac{x (4 + f^{2})}{(1 + x) (1 - x)}$

Step 1.5.3

Cancel the common factor of $4 + f^{2}$ .

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Step 1.5.3.1

Factor $4 + f^{2}$ out of $x (4 + f^{2})$ .

$\frac{f}{4 + f^{2}} \frac{d f}{d x} = \frac{1}{4 + f^{2}} \cdot \frac{(4 + f^{2}) x}{(1 + x) (1 - x)}$

Step 1.5.3.2

Cancel the common factor.

$\frac{f}{4 + f^{2}} \frac{d f}{d x} = \frac{1}{4 + f^{2}} \cdot \frac{(4 + f^{2}) x}{(1 + x) (1 - x)}$

Step 1.5.3.3

Rewrite the expression.

$\frac{f}{4 + f^{2}} \frac{d f}{d x} = \frac{x}{(1 + x) (1 - x)}$

Step 1.6

Rewrite the equation.

$\frac{f}{4 + f^{2}} d f = \frac{x}{(1 + x) (1 - x)} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{f}{4 + f^{2}} d f = \int \frac{x}{(1 + x) (1 - x)} d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u_{1} = 4 + f^{2}$ . Then $d u_{1} = 2 f d f$ , so $\frac{1}{2} d u_{1} = f d f$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 2.2.1.1

Let $u_{1} = 4 + f^{2}$ . Find $\frac{d u_{1}}{d f}$ .

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Step 2.2.1.1.1

Differentiate $4 + f^{2}$ .

$\frac{d}{d f} [4 + f^{2}]$

Step 2.2.1.1.2

By the Sum Rule, the derivative of $4 + f^{2}$ with respect to $f$ is $\frac{d}{d f} [4] + \frac{d}{d f} [f^{2}]$ .

$\frac{d}{d f} [4] + \frac{d}{d f} [f^{2}]$

Step 2.2.1.1.3

Since $4$ is constant with respect to $f$ , the derivative of $4$ with respect to $f$ is $0$ .

$0 + \frac{d}{d f} [f^{2}]$

Step 2.2.1.1.4

Differentiate using the Power Rule which states that $\frac{d}{d f} [f^{n}]$ is $n f^{n - 1}$ where $n = 2$ .

$0 + 2 f$

Step 2.2.1.1.5

Add $0$ and $2 f$ .

$2 f$

Step 2.2.1.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\int \frac{1}{u_{1}} \cdot \frac{1}{2} d u_{1} = \int \frac{x}{(1 + x) (1 - x)} d x$

Step 2.2.2

Simplify.

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Step 2.2.2.1

Multiply $\frac{1}{u_{1}}$ by $\frac{1}{2}$ .

$\int \frac{1}{u_{1} \cdot 2} d u_{1} = \int \frac{x}{(1 + x) (1 - x)} d x$

Step 2.2.2.2

Move $2$ to the left of $u_{1}$ .

$\int \frac{1}{2 u_{1}} d u_{1} = \int \frac{x}{(1 + x) (1 - x)} d x$

Step 2.2.3

Since $\frac{1}{2}$ is constant with respect to $u_{1}$ , move $\frac{1}{2}$ out of the integral.

$\frac{1}{2} \int \frac{1}{u_{1}} d u_{1} = \int \frac{x}{(1 + x) (1 - x)} d x$

Step 2.2.4

The integral of $\frac{1}{u_{1}}$ with respect to $u_{1}$ is $\ln (| u_{1} |)$ .

$\frac{1}{2} (\ln (| u_{1} |) + C_{1}) = \int \frac{x}{(1 + x) (1 - x)} d x$

Step 2.2.5

Simplify.

$\frac{1}{2} \ln (| u_{1} |) + C_{1} = \int \frac{x}{(1 + x) (1 - x)} d x$

Step 2.2.6

Replace all occurrences of $u_{1}$ with $4 + f^{2}$ .

$\frac{1}{2} \ln (| 4 + f^{2} |) + C_{1} = \int \frac{x}{(1 + x) (1 - x)} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Let $u_{2} = (1 + x) (1 - x)$ . Then $d u_{2} = - 2 x d x$ , so $- \frac{1}{2} d u_{2} = x d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 2.3.1.1

Let $u_{2} = (1 + x) (1 - x)$ . Find $\frac{d u_{2}}{d x}$ .

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Step 2.3.1.1.1

Differentiate $(1 + x) (1 - x)$ .

$\frac{d}{d x} [(1 + x) (1 - x)]$

Step 2.3.1.1.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = 1 + x$ and $g (x) = 1 - x$ .

$(1 + x) \frac{d}{d x} [1 - x] + (1 - x) \frac{d}{d x} [1 + x]$

Step 2.3.1.1.3

Differentiate.

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Step 2.3.1.1.3.1

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

$(1 + x) (\frac{d}{d x} [1] + \frac{d}{d x} [- x]) + (1 - x) \frac{d}{d x} [1 + x]$

Step 2.3.1.1.3.2

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

$(1 + x) (0 + \frac{d}{d x} [- x]) + (1 - x) \frac{d}{d x} [1 + x]$

Step 2.3.1.1.3.3

Add $0$ and $\frac{d}{d x} [- x]$ .

$(1 + x) \frac{d}{d x} [- x] + (1 - x) \frac{d}{d x} [1 + x]$

Step 2.3.1.1.3.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$(1 + x) (- \frac{d}{d x} [x]) + (1 - x) \frac{d}{d x} [1 + x]$

Step 2.3.1.1.3.5

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

$(1 + x) (- 1 \cdot 1) + (1 - x) \frac{d}{d x} [1 + x]$

Step 2.3.1.1.3.6

Simplify the expression.

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Step 2.3.1.1.3.6.1

Multiply $- 1$ by $1$ .

$(1 + x) \cdot - 1 + (1 - x) \frac{d}{d x} [1 + x]$

Step 2.3.1.1.3.6.2

Move $- 1$ to the left of $1 + x$ .

$- 1 \cdot (1 + x) + (1 - x) \frac{d}{d x} [1 + x]$

Step 2.3.1.1.3.6.3

Rewrite $- 1 (1 + x)$ as $- (1 + x)$ .

$- (1 + x) + (1 - x) \frac{d}{d x} [1 + x]$

Step 2.3.1.1.3.7

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

$- (1 + x) + (1 - x) (\frac{d}{d x} [1] + \frac{d}{d x} [x])$

Step 2.3.1.1.3.8

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

$- (1 + x) + (1 - x) (0 + \frac{d}{d x} [x])$

Step 2.3.1.1.3.9

Add $0$ and $\frac{d}{d x} [x]$ .

$- (1 + x) + (1 - x) \frac{d}{d x} [x]$

Step 2.3.1.1.3.10

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

$- (1 + x) + (1 - x) \cdot 1$

Step 2.3.1.1.3.11

Multiply $1 - x$ by $1$ .

$- (1 + x) + 1 - x$

Step 2.3.1.1.4

Simplify.

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Step 2.3.1.1.4.1

Apply the distributive property.

$- 1 \cdot 1 - x + 1 - x$

Step 2.3.1.1.4.2

Combine terms.

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Step 2.3.1.1.4.2.1

Multiply $- 1$ by $1$ .

$- 1 - x + 1 - x$

Step 2.3.1.1.4.2.2

Add $- 1$ and $1$ .

$- x + 0 - x$

Step 2.3.1.1.4.2.3

Add $- x$ and $0$ .

$- x - x$

Step 2.3.1.1.4.2.4

Subtract $x$ from $- x$ .

$- 2 x$

Step 2.3.1.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\frac{1}{2} \ln (| 4 + f^{2} |) + C_{1} = \int \frac{1}{u_{2}} \cdot \frac{1}{- 2} d u_{2}$

Step 2.3.2

Simplify.

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Step 2.3.2.1

Move the negative in front of the fraction.

$\frac{1}{2} \ln (| 4 + f^{2} |) + C_{1} = \int \frac{1}{u_{2}} (- \frac{1}{2}) d u_{2}$

Step 2.3.2.2

Multiply $\frac{1}{u_{2}}$ by $\frac{1}{2}$ .

$\frac{1}{2} \ln (| 4 + f^{2} |) + C_{1} = \int - \frac{1}{u_{2} \cdot 2} d u_{2}$

Step 2.3.2.3

Move $2$ to the left of $u_{2}$ .

$\frac{1}{2} \ln (| 4 + f^{2} |) + C_{1} = \int - \frac{1}{2 u_{2}} d u_{2}$

Step 2.3.3

Since $- 1$ is constant with respect to $u_{2}$ , move $- 1$ out of the integral.

$\frac{1}{2} \ln (| 4 + f^{2} |) + C_{1} = - \int \frac{1}{2 u_{2}} d u_{2}$

Step 2.3.4

Since $\frac{1}{2}$ is constant with respect to $u_{2}$ , move $\frac{1}{2}$ out of the integral.

$\frac{1}{2} \ln (| 4 + f^{2} |) + C_{1} = - (\frac{1}{2} \int \frac{1}{u_{2}} d u_{2})$

Step 2.3.5

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$\frac{1}{2} \ln (| 4 + f^{2} |) + C_{1} = - \frac{1}{2} (\ln (| u_{2} |) + C_{2})$

Step 2.3.6

Simplify.

$\frac{1}{2} \ln (| 4 + f^{2} |) + C_{1} = - \frac{1}{2} \ln (| u_{2} |) + C_{2}$

Step 2.3.7

Replace all occurrences of $u_{2}$ with $(1 + x) (1 - x)$ .

$\frac{1}{2} \ln (| 4 + f^{2} |) + C_{1} = - \frac{1}{2} \ln (| (1 + x) (1 - x) |) + C_{2}$

Step 2.4

Group the constant of integration on the right side as $C$ .

$\frac{1}{2} \ln (| 4 + f^{2} |) = - \frac{1}{2} \ln (| (1 + x) (1 - x) |) + C$

Step 3

Solve for $f$ .

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Step 3.1

Multiply both sides of the equation by $2$ .

$2 (\frac{1}{2} \ln (| 4 + f^{2} |)) = 2 (- \frac{1}{2} \ln (| (1 + x) (1 - x) |) + C)$

Step 3.2

Simplify both sides of the equation.

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

Simplify $2 (\frac{1}{2} \ln (| 4 + f^{2} |))$ .

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Step 3.2.1.1.1

Combine $\frac{1}{2}$ and $\ln (| 4 + f^{2} |)$ .

$2 \frac{\ln (| 4 + f^{2} |)}{2} = 2 (- \frac{1}{2} \ln (| (1 + x) (1 - x) |) + C)$

Step 3.2.1.1.2

Cancel the common factor of $2$ .

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Step 3.2.1.1.2.1

Cancel the common factor.

$2 \frac{\ln (| 4 + f^{2} |)}{2} = 2 (- \frac{1}{2} \ln (| (1 + x) (1 - x) |) + C)$

Step 3.2.1.1.2.2

Rewrite the expression.

$\ln (| 4 + f^{2} |) = 2 (- \frac{1}{2} \ln (| (1 + x) (1 - x) |) + C)$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $2 (- \frac{1}{2} \ln (| (1 + x) (1 - x) |) + C)$ .

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Step 3.2.2.1.1

Simplify each term.

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Step 3.2.2.1.1.1

Expand $(1 + x) (1 - x)$ using the FOIL Method.

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Step 3.2.2.1.1.1.1

Apply the distributive property.

$\ln (| 4 + f^{2} |) = 2 (- \frac{1}{2} \ln (| 1 (1 - x) + x (1 - x) |) + C)$

Step 3.2.2.1.1.1.2

Apply the distributive property.

$\ln (| 4 + f^{2} |) = 2 (- \frac{1}{2} \ln (| 1 \cdot 1 + 1 (- x) + x (1 - x) |) + C)$

Step 3.2.2.1.1.1.3

Apply the distributive property.

$\ln (| 4 + f^{2} |) = 2 (- \frac{1}{2} \ln (| 1 \cdot 1 + 1 (- x) + x \cdot 1 + x (- x) |) + C)$

Step 3.2.2.1.1.2

Simplify and combine like terms.

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Step 3.2.2.1.1.2.1

Simplify each term.

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Step 3.2.2.1.1.2.1.1

Multiply $1$ by $1$ .

$\ln (| 4 + f^{2} |) = 2 (- \frac{1}{2} \ln (| 1 + 1 (- x) + x \cdot 1 + x (- x) |) + C)$

Step 3.2.2.1.1.2.1.2

Multiply $- x$ by $1$ .

$\ln (| 4 + f^{2} |) = 2 (- \frac{1}{2} \ln (| 1 - x + x \cdot 1 + x (- x) |) + C)$

Step 3.2.2.1.1.2.1.3

Multiply $x$ by $1$ .

$\ln (| 4 + f^{2} |) = 2 (- \frac{1}{2} \ln (| 1 - x + x + x (- x) |) + C)$

Step 3.2.2.1.1.2.1.4

Rewrite using the commutative property of multiplication.

$\ln (| 4 + f^{2} |) = 2 (- \frac{1}{2} \ln (| 1 - x + x - x \cdot x |) + C)$

Step 3.2.2.1.1.2.1.5

Multiply $x$ by $x$ by adding the exponents.

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Step 3.2.2.1.1.2.1.5.1

Move $x$ .

$\ln (| 4 + f^{2} |) = 2 (- \frac{1}{2} \ln (| 1 - x + x - (x \cdot x) |) + C)$

Step 3.2.2.1.1.2.1.5.2

Multiply $x$ by $x$ .

$\ln (| 4 + f^{2} |) = 2 (- \frac{1}{2} \ln (| 1 - x + x - x^{2} |) + C)$

Step 3.2.2.1.1.2.2

Add $- x$ and $x$ .

$\ln (| 4 + f^{2} |) = 2 (- \frac{1}{2} \ln (| 1 + 0 - x^{2} |) + C)$

Step 3.2.2.1.1.2.3

Add $1$ and $0$ .

$\ln (| 4 + f^{2} |) = 2 (- \frac{1}{2} \ln (| 1 - x^{2} |) + C)$

Step 3.2.2.1.1.3

Combine $\ln (| 1 - x^{2} |)$ and $\frac{1}{2}$ .

$\ln (| 4 + f^{2} |) = 2 (- \frac{\ln (| 1 - x^{2} |)}{2} + C)$

Step 3.2.2.1.2

To write $C$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\ln (| 4 + f^{2} |) = 2 (- \frac{\ln (| 1 - x^{2} |)}{2} + C \cdot \frac{2}{2})$

Step 3.2.2.1.3

Simplify terms.

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Step 3.2.2.1.3.1

Combine $C$ and $\frac{2}{2}$ .

$\ln (| 4 + f^{2} |) = 2 (- \frac{\ln (| 1 - x^{2} |)}{2} + \frac{C \cdot 2}{2})$

Step 3.2.2.1.3.2

Combine the numerators over the common denominator.

$\ln (| 4 + f^{2} |) = 2 \frac{- \ln (| 1 - x^{2} |) + C \cdot 2}{2}$

Step 3.2.2.1.3.3

Cancel the common factor of $2$ .

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Step 3.2.2.1.3.3.1

Cancel the common factor.

$\ln (| 4 + f^{2} |) = 2 \frac{- \ln (| 1 - x^{2} |) + C \cdot 2}{2}$

Step 3.2.2.1.3.3.2

Rewrite the expression.

$\ln (| 4 + f^{2} |) = - \ln (| 1 - x^{2} |) + C \cdot 2$

Step 3.2.2.1.4

Move $2$ to the left of $C$ .

$\ln (| 4 + f^{2} |) = - \ln (| 1 - x^{2} |) + 2 C$

Step 3.3

Move all the terms containing a logarithm to the left side of the equation.

$\ln (| 4 + f^{2} |) + \ln (| 1 - x^{2} |) = 2 C$

Step 3.4

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\ln (| 4 + f^{2} | | 1 - x^{2} |) = 2 C$

Step 3.5

To multiply absolute values, multiply the terms inside each absolute value.

$\ln (| (4 + f^{2}) (1 - x^{2}) |) = 2 C$

Step 3.6

Expand $(4 + f^{2}) (1 - x^{2})$ using the FOIL Method.

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Step 3.6.1

Apply the distributive property.

$\ln (| 4 (1 - x^{2}) + f^{2} (1 - x^{2}) |) = 2 C$

Step 3.6.2

Apply the distributive property.

$\ln (| 4 \cdot 1 + 4 (- x^{2}) + f^{2} (1 - x^{2}) |) = 2 C$

Step 3.6.3

Apply the distributive property.

$\ln (| 4 \cdot 1 + 4 (- x^{2}) + f^{2} \cdot 1 + f^{2} (- x^{2}) |) = 2 C$

Step 3.7

Simplify each term.

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Step 3.7.1

Multiply $4$ by $1$ .

$\ln (| 4 + 4 (- x^{2}) + f^{2} \cdot 1 + f^{2} (- x^{2}) |) = 2 C$

Step 3.7.2

Multiply $- 1$ by $4$ .

$\ln (| 4 - 4 x^{2} + f^{2} \cdot 1 + f^{2} (- x^{2}) |) = 2 C$

Step 3.7.3

Multiply $f^{2}$ by $1$ .

$\ln (| 4 - 4 x^{2} + f^{2} + f^{2} (- x^{2}) |) = 2 C$

Step 3.7.4

Rewrite using the commutative property of multiplication.

$\ln (| 4 - 4 x^{2} + f^{2} - f^{2} x^{2} |) = 2 C$

Step 3.8

To solve for $f$ , rewrite the equation using properties of logarithms.

$e^{\ln (| 4 - 4 x^{2} + f^{2} - f^{2} x^{2} |)} = e^{2 C}$

Step 3.9

Rewrite $\ln (| 4 - 4 x^{2} + f^{2} - f^{2} x^{2} |) = 2 C$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{2 C} = | 4 - 4 x^{2} + f^{2} - f^{2} x^{2} |$

Step 3.10

Solve for $f$ .

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Step 3.10.1

Rewrite the equation as $| 4 - 4 x^{2} + f^{2} - f^{2} x^{2} | = e^{2 C}$ .

$| 4 - 4 x^{2} + f^{2} - f^{2} x^{2} | = e^{2 C}$

Step 3.10.2

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$4 - 4 x^{2} + f^{2} - f^{2} x^{2} = \pm e^{2 C}$

Step 3.10.3

Move all terms not containing $f$ to the right side of the equation.

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Step 3.10.3.1

Subtract $4$ from both sides of the equation.

$- 4 x^{2} + f^{2} - f^{2} x^{2} = \pm e^{2 C} - 4$

Step 3.10.3.2

Add $4 x^{2}$ to both sides of the equation.

$f^{2} - f^{2} x^{2} = \pm e^{2 C} - 4 + 4 x^{2}$

Step 3.10.4

Factor $f^{2}$ out of $f^{2} - f^{2} x^{2}$ .

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Step 3.10.4.1

Multiply by $1$ .

$f^{2} \cdot 1 - f^{2} x^{2} = \pm e^{2 C} - 4 + 4 x^{2}$

Step 3.10.4.2

Factor $f^{2}$ out of $- f^{2} x^{2}$ .

$f^{2} \cdot 1 + f^{2} (- 1 x^{2}) = \pm e^{2 C} - 4 + 4 x^{2}$

Step 3.10.4.3

Factor $f^{2}$ out of $f^{2} \cdot 1 + f^{2} (- 1 x^{2})$ .

$f^{2} (1 - 1 x^{2}) = \pm e^{2 C} - 4 + 4 x^{2}$

Step 3.10.5

Rewrite $1$ as $1^{2}$ .

$f^{2} (1^{2} - 1 x^{2}) = \pm e^{2 C} - 4 + 4 x^{2}$

Step 3.10.6

Factor.

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Step 3.10.6.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = x$ .

$f^{2} ((1 + x) (1 - x)) = \pm e^{2 C} - 4 + 4 x^{2}$

Step 3.10.6.2

Remove unnecessary parentheses.

$f^{2} (1 + x) (1 - x) = \pm e^{2 C} - 4 + 4 x^{2}$

Step 3.10.7

Divide each term in $f^{2} (1 + x) (1 - x) = \pm e^{2 C} - 4 + 4 x^{2}$ by $(1 + x) (1 - x)$ and simplify.

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Step 3.10.7.1

Divide each term in $f^{2} (1 + x) (1 - x) = \pm e^{2 C} - 4 + 4 x^{2}$ by $(1 + x) (1 - x)$ .

$\frac{f^{2} (1 + x) (1 - x)}{(1 + x) (1 - x)} = \frac{\pm e^{2 C}}{(1 + x) (1 - x)} + \frac{- 4}{(1 + x) (1 - x)} + \frac{4 x^{2}}{(1 + x) (1 - x)}$

Step 3.10.7.2

Simplify the left side.

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Step 3.10.7.2.1

Cancel the common factor of $1 + x$ .

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Step 3.10.7.2.1.1

Cancel the common factor.

$\frac{f^{2} (1 + x) (1 - x)}{(1 + x) (1 - x)} = \frac{\pm e^{2 C}}{(1 + x) (1 - x)} + \frac{- 4}{(1 + x) (1 - x)} + \frac{4 x^{2}}{(1 + x) (1 - x)}$

Step 3.10.7.2.1.2

Rewrite the expression.

$\frac{f^{2} (1 - x)}{1 - x} = \frac{\pm e^{2 C}}{(1 + x) (1 - x)} + \frac{- 4}{(1 + x) (1 - x)} + \frac{4 x^{2}}{(1 + x) (1 - x)}$

Step 3.10.7.2.2

Cancel the common factor of $1 - x$ .

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Step 3.10.7.2.2.1

Cancel the common factor.

$\frac{f^{2} (1 - x)}{1 - x} = \frac{\pm e^{2 C}}{(1 + x) (1 - x)} + \frac{- 4}{(1 + x) (1 - x)} + \frac{4 x^{2}}{(1 + x) (1 - x)}$

Step 3.10.7.2.2.2

Divide $f^{2}$ by $1$ .

$f^{2} = \frac{\pm e^{2 C}}{(1 + x) (1 - x)} + \frac{- 4}{(1 + x) (1 - x)} + \frac{4 x^{2}}{(1 + x) (1 - x)}$

Step 3.10.7.3

Simplify the right side.

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Step 3.10.7.3.1

Move the negative in front of the fraction.

$f^{2} = \frac{\pm e^{2 C}}{(1 + x) (1 - x)} - \frac{4}{(1 + x) (1 - x)} + \frac{4 x^{2}}{(1 + x) (1 - x)}$

Step 3.10.8

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$f = \pm \sqrt{\frac{\pm e^{2 C}}{(1 + x) (1 - x)} - \frac{4}{(1 + x) (1 - x)} + \frac{4 x^{2}}{(1 + x) (1 - x)}}$

Step 3.10.9

Simplify $\pm \sqrt{\frac{\pm e^{2 C}}{(1 + x) (1 - x)} - \frac{4}{(1 + x) (1 - x)} + \frac{4 x^{2}}{(1 + x) (1 - x)}}$ .

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Step 3.10.9.1

Combine the numerators over the common denominator.

$f = \pm \sqrt{\frac{\pm e^{2 C} - 4}{(1 + x) (1 - x)} + \frac{4 x^{2}}{(1 + x) (1 - x)}}$

Step 3.10.9.2

Combine the numerators over the common denominator.

$f = \pm \sqrt{\frac{\pm e^{2 C} - 4 + 4 x^{2}}{(1 + x) (1 - x)}}$

Step 3.10.9.3

Rewrite $\sqrt{\frac{\pm e^{2 C} - 4 + 4 x^{2}}{(1 + x) (1 - x)}}$ as $\frac{\sqrt{\pm e^{2 C} - 4 + 4 x^{2}}}{\sqrt{(1 + x) (1 - x)}}$ .

$f = \pm \frac{\sqrt{\pm e^{2 C} - 4 + 4 x^{2}}}{\sqrt{(1 + x) (1 - x)}}$

Step 3.10.9.4

Multiply $\frac{\sqrt{\pm e^{2 C} - 4 + 4 x^{2}}}{\sqrt{(1 + x) (1 - x)}}$ by $\frac{\sqrt{(1 + x) (1 - x)}}{\sqrt{(1 + x) (1 - x)}}$ .

$f = \pm \frac{\sqrt{\pm e^{2 C} - 4 + 4 x^{2}}}{\sqrt{(1 + x) (1 - x)}} \cdot \frac{\sqrt{(1 + x) (1 - x)}}{\sqrt{(1 + x) (1 - x)}}$

Step 3.10.9.5

Combine and simplify the denominator.

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Step 3.10.9.5.1

Multiply $\frac{\sqrt{\pm e^{2 C} - 4 + 4 x^{2}}}{\sqrt{(1 + x) (1 - x)}}$ by $\frac{\sqrt{(1 + x) (1 - x)}}{\sqrt{(1 + x) (1 - x)}}$ .

$f = \pm \frac{\sqrt{\pm e^{2 C} - 4 + 4 x^{2}} \sqrt{(1 + x) (1 - x)}}{\sqrt{(1 + x) (1 - x)} \sqrt{(1 + x) (1 - x)}}$

Step 3.10.9.5.2

Raise $\sqrt{(1 + x) (1 - x)}$ to the power of $1$ .

$f = \pm \frac{\sqrt{\pm e^{2 C} - 4 + 4 x^{2}} \sqrt{(1 + x) (1 - x)}}{{\sqrt{(1 + x) (1 - x)}}^{1} \sqrt{(1 + x) (1 - x)}}$

Step 3.10.9.5.3

Raise $\sqrt{(1 + x) (1 - x)}$ to the power of $1$ .

$f = \pm \frac{\sqrt{\pm e^{2 C} - 4 + 4 x^{2}} \sqrt{(1 + x) (1 - x)}}{{\sqrt{(1 + x) (1 - x)}}^{1} {\sqrt{(1 + x) (1 - x)}}^{1}}$

Step 3.10.9.5.4

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

$f = \pm \frac{\sqrt{\pm e^{2 C} - 4 + 4 x^{2}} \sqrt{(1 + x) (1 - x)}}{{\sqrt{(1 + x) (1 - x)}}^{1 + 1}}$

Step 3.10.9.5.5

Add $1$ and $1$ .

$f = \pm \frac{\sqrt{\pm e^{2 C} - 4 + 4 x^{2}} \sqrt{(1 + x) (1 - x)}}{{\sqrt{(1 + x) (1 - x)}}^{2}}$

Step 3.10.9.5.6

Rewrite ${\sqrt{(1 + x) (1 - x)}}^{2}$ as $(1 + x) (1 - x)$ .

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Step 3.10.9.5.6.1

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

$f = \pm \frac{\sqrt{\pm e^{2 C} - 4 + 4 x^{2}} \sqrt{(1 + x) (1 - x)}}{{({((1 + x) (1 - x))}^{\frac{1}{2}})}^{2}}$

Step 3.10.9.5.6.2

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

$f = \pm \frac{\sqrt{\pm e^{2 C} - 4 + 4 x^{2}} \sqrt{(1 + x) (1 - x)}}{{((1 + x) (1 - x))}^{\frac{1}{2} \cdot 2}}$

Step 3.10.9.5.6.3

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

$f = \pm \frac{\sqrt{\pm e^{2 C} - 4 + 4 x^{2}} \sqrt{(1 + x) (1 - x)}}{{((1 + x) (1 - x))}^{\frac{2}{2}}}$

Step 3.10.9.5.6.4

Cancel the common factor of $2$ .

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Step 3.10.9.5.6.4.1

Cancel the common factor.

$f = \pm \frac{\sqrt{\pm e^{2 C} - 4 + 4 x^{2}} \sqrt{(1 + x) (1 - x)}}{{((1 + x) (1 - x))}^{\frac{2}{2}}}$

Step 3.10.9.5.6.4.2

Rewrite the expression.

$f = \pm \frac{\sqrt{\pm e^{2 C} - 4 + 4 x^{2}} \sqrt{(1 + x) (1 - x)}}{{((1 + x) (1 - x))}^{1}}$

Step 3.10.9.5.6.5

Simplify.

$f = \pm \frac{\sqrt{\pm e^{2 C} - 4 + 4 x^{2}} \sqrt{(1 + x) (1 - x)}}{(1 + x) (1 - x)}$

Step 3.10.9.6

Combine using the product rule for radicals.

$f = \pm \frac{\sqrt{(\pm e^{2 C} - 4 + 4 x^{2}) (1 + x) (1 - x)}}{(1 + x) (1 - x)}$

Step 4

Simplify the constant of integration.

$f = \pm \frac{\sqrt{(C + 4 x^{2}) (1 + x) (1 - x)}}{(1 + x) (1 - x)}$