Calculus Examples

$\frac{d x}{d t} = (x - 1) (1 - 2 x)$

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{(x - 1) (1 - 2 x)}$ .

$\frac{1}{(x - 1) (1 - 2 x)} \frac{d x}{d t} = \frac{1}{(x - 1) (1 - 2 x)} ((x - 1) (1 - 2 x))$

Step 1.2

Cancel the common factor of $(x - 1) (1 - 2 x)$ .

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Step 1.2.1

Cancel the common factor.

$\frac{1}{(x - 1) (1 - 2 x)} \frac{d x}{d t} = \frac{1}{(x - 1) (1 - 2 x)} ((x - 1) (1 - 2 x))$

Step 1.2.2

Rewrite the expression.

$\frac{1}{(x - 1) (1 - 2 x)} \frac{d x}{d t} = 1$

Step 1.3

Rewrite the equation.

$\frac{1}{(x - 1) (1 - 2 x)} d x = 1 d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Write the fraction using partial fraction decomposition.

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Step 2.2.1.1

Decompose the fraction and multiply through by the common denominator.

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Step 2.2.1.1.1

For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor in the denominator is linear, put a single variable in its place $A$ .

$\frac{A}{x - 1}$

Step 2.2.1.1.2

$\frac{A}{x - 1} + \frac{B}{1 - 2 x}$

Step 2.2.1.1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(x - 1) (1 - 2 x)$ .

$\frac{1 (x - 1) (1 - 2 x)}{(x - 1) (1 - 2 x)} = \frac{(A) (x - 1) (1 - 2 x)}{x - 1} + \frac{(B) (x - 1) (1 - 2 x)}{1 - 2 x}$

Step 2.2.1.1.4

Cancel the common factor of $x - 1$ .

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Step 2.2.1.1.4.1

Cancel the common factor.

$\frac{1 (x - 1) (1 - 2 x)}{(x - 1) (1 - 2 x)} = \frac{(A) (x - 1) (1 - 2 x)}{x - 1} + \frac{(B) (x - 1) (1 - 2 x)}{1 - 2 x}$

Step 2.2.1.1.4.2

Rewrite the expression.

$\frac{1 (1 - 2 x)}{1 - 2 x} = \frac{(A) (x - 1) (1 - 2 x)}{x - 1} + \frac{(B) (x - 1) (1 - 2 x)}{1 - 2 x}$

Step 2.2.1.1.5

Cancel the common factor of $1 - 2 x$ .

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Step 2.2.1.1.5.1

Cancel the common factor.

$\frac{1 (1 - 2 x)}{1 - 2 x} = \frac{(A) (x - 1) (1 - 2 x)}{x - 1} + \frac{(B) (x - 1) (1 - 2 x)}{1 - 2 x}$

Step 2.2.1.1.5.2

Rewrite the expression.

$1 = \frac{(A) (x - 1) (1 - 2 x)}{x - 1} + \frac{(B) (x - 1) (1 - 2 x)}{1 - 2 x}$

Step 2.2.1.1.6

Simplify each term.

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Step 2.2.1.1.6.1

Cancel the common factor of $x - 1$ .

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Step 2.2.1.1.6.1.1

Cancel the common factor.

$1 = \frac{A (x - 1) (1 - 2 x)}{x - 1} + \frac{(B) (x - 1) (1 - 2 x)}{1 - 2 x}$

Step 2.2.1.1.6.1.2

Divide $(A) (1 - 2 x)$ by $1$ .

$1 = (A) (1 - 2 x) + \frac{(B) (x - 1) (1 - 2 x)}{1 - 2 x}$

Step 2.2.1.1.6.2

Apply the distributive property.

$1 = A \cdot 1 + A (- 2 x) + \frac{(B) (x - 1) (1 - 2 x)}{1 - 2 x}$

Step 2.2.1.1.6.3

Multiply $A$ by $1$ .

$1 = A + A (- 2 x) + \frac{(B) (x - 1) (1 - 2 x)}{1 - 2 x}$

Step 2.2.1.1.6.4

Rewrite using the commutative property of multiplication.

$1 = A - 2 A x + \frac{(B) (x - 1) (1 - 2 x)}{1 - 2 x}$

Step 2.2.1.1.6.5

Cancel the common factor of $1 - 2 x$ .

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Step 2.2.1.1.6.5.1

Cancel the common factor.

$1 = A - 2 A x + \frac{(B) (x - 1) (1 - 2 x)}{1 - 2 x}$

Step 2.2.1.1.6.5.2

Divide $(B) (x - 1)$ by $1$ .

$1 = A - 2 A x + (B) (x - 1)$

Step 2.2.1.1.6.6

Apply the distributive property.

$1 = A - 2 A x + B x + B \cdot - 1$

Step 2.2.1.1.6.7

Move $- 1$ to the left of $B$ .

$1 = A - 2 A x + B x - 1 \cdot B$

Step 2.2.1.1.6.8

Rewrite $- 1 B$ as $- B$ .

$1 = A - 2 A x + B x - B$

Step 2.2.1.1.7

Simplify the expression.

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Step 2.2.1.1.7.1

Move $A$ .

$1 = A - 2 x A + B x - B$

Step 2.2.1.1.7.2

Reorder $B$ and $x$ .

$1 = A - 2 x A + B x - B$

Step 2.2.1.1.7.3

Move $A$ .

$1 = - 2 x A + B x + A - B$

Step 2.2.1.2

Create equations for the partial fraction variables and use them to set up a system of equations.

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Step 2.2.1.2.1

Create an equation for the partial fraction variables by equating the coefficients of $x$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = - 2 A + B$

Step 2.2.1.2.2

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing $x$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$1 = A - 1 B$

Step 2.2.1.2.3

Set up the system of equations to find the coefficients of the partial fractions.

$0 = - 2 A + B$

$1 = A - 1 B$

$0 = - 2 A + B$

$1 = A - 1 B$

Step 2.2.1.3

Solve the system of equations.

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Step 2.2.1.3.1

Solve for $B$ in $0 = - 2 A + B$ .

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Step 2.2.1.3.1.1

Rewrite the equation as $- 2 A + B = 0$ .

$- 2 A + B = 0$

$1 = A - 1 B$

Step 2.2.1.3.1.2

Add $2 A$ to both sides of the equation.

$B = 2 A$

$1 = A - 1 B$

$B = 2 A$

$1 = A - 1 B$

Step 2.2.1.3.2

Replace all occurrences of $B$ with $2 A$ in each equation.

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Step 2.2.1.3.2.1

Replace all occurrences of $B$ in $1 = A - 1 B$ with $2 A$ .

$1 = A - 1 (2 A)$

$B = 2 A$

Step 2.2.1.3.2.2

Simplify the right side.

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Step 2.2.1.3.2.2.1

Simplify $A - 1 (2 A)$ .

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Step 2.2.1.3.2.2.1.1

Multiply $2$ by $- 1$ .

$1 = A - 2 A$

$B = 2 A$

Step 2.2.1.3.2.2.1.2

Subtract $2 A$ from $A$ .

$1 = - A$

$B = 2 A$

$1 = - A$

$B = 2 A$

$1 = - A$

$B = 2 A$

$1 = - A$

$B = 2 A$

Step 2.2.1.3.3

Solve for $A$ in $1 = - A$ .

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Step 2.2.1.3.3.1

Rewrite the equation as $- A = 1$ .

$- A = 1$

$B = 2 A$

Step 2.2.1.3.3.2

Divide each term in $- A = 1$ by $- 1$ and simplify.

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Step 2.2.1.3.3.2.1

Divide each term in $- A = 1$ by $- 1$ .

$\frac{- A}{- 1} = \frac{1}{- 1}$

$B = 2 A$

Step 2.2.1.3.3.2.2

Simplify the left side.

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Step 2.2.1.3.3.2.2.1

Dividing two negative values results in a positive value.

$\frac{A}{1} = \frac{1}{- 1}$

$B = 2 A$

Step 2.2.1.3.3.2.2.2

Divide $A$ by $1$ .

$A = \frac{1}{- 1}$

$B = 2 A$

$A = \frac{1}{- 1}$

$B = 2 A$

Step 2.2.1.3.3.2.3

Simplify the right side.

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Step 2.2.1.3.3.2.3.1

Divide $1$ by $- 1$ .

$A = - 1$

$B = 2 A$

$A = - 1$

$B = 2 A$

$A = - 1$

$B = 2 A$

$A = - 1$

$B = 2 A$

Step 2.2.1.3.4

Replace all occurrences of $A$ with $- 1$ in each equation.

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Step 2.2.1.3.4.1

Replace all occurrences of $A$ in $B = 2 A$ with $- 1$ .

$B = 2 (- 1)$

$A = - 1$

Step 2.2.1.3.4.2

Simplify the right side.

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Step 2.2.1.3.4.2.1

Multiply $2$ by $- 1$ .

$B = - 2$

$A = - 1$

$B = - 2$

$A = - 1$

$B = - 2$

$A = - 1$

Step 2.2.1.3.5

List all of the solutions.

$B = - 2, A = - 1$

Step 2.2.1.4

Replace each of the partial fraction coefficients in $\frac{A}{x - 1} + \frac{B}{1 - 2 x}$ with the values found for $A$ and $B$ .

$\frac{- 1}{x - 1} + \frac{- 2}{1 - 2 x}$

Step 2.2.1.5

Simplify.

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Step 2.2.1.5.1

Move the negative in front of the fraction.

$- \frac{1}{x - 1} + \frac{- 2}{1 - 2 x}$

Step 2.2.1.5.2

Move the negative in front of the fraction.

$\int - \frac{1}{x - 1} - \frac{2}{1 - 2 x} d x = \int d t$

Step 2.2.2

Split the single integral into multiple integrals.

$\int - \frac{1}{x - 1} d x + \int - \frac{2}{1 - 2 x} d x = \int d t$

Step 2.2.3

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

$- \int \frac{1}{x - 1} d x + \int - \frac{2}{1 - 2 x} d x = \int d t$

Step 2.2.4

Let $u_{1} = x - 1$ . Then $d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 2.2.4.1

Let $u_{1} = x - 1$ . Find $\frac{d u_{1}}{d x}$ .

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Step 2.2.4.1.1

Differentiate $x - 1$ .

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

Step 2.2.4.1.2

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 2.2.4.1.3

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 2.2.4.1.4

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

$1 + 0$

Step 2.2.4.1.5

Add $1$ and $0$ .

$1$

Step 2.2.4.2

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

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

Step 2.2.5

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

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

Step 2.2.6

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

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

Step 2.2.7

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

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

Step 2.2.8

Multiply $2$ by $- 1$ .

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

Step 2.2.9

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

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Step 2.2.9.1

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

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Step 2.2.9.1.1

Differentiate $1 - 2 x$ .

$\frac{d}{d x} [1 - 2 x]$

Step 2.2.9.1.2

Differentiate.

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Step 2.2.9.1.2.1

By the Sum Rule, the derivative of $1 - 2 x$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- 2 x]$ .

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

Step 2.2.9.1.2.2

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

$0 + \frac{d}{d x} [- 2 x]$

Step 2.2.9.1.3

Evaluate $\frac{d}{d x} [- 2 x]$ .

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Step 2.2.9.1.3.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x$ with respect to $x$ is $- 2 \frac{d}{d x} [x]$ .

$0 - 2 \frac{d}{d x} [x]$

Step 2.2.9.1.3.2

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

$0 - 2 \cdot 1$

Step 2.2.9.1.3.3

Multiply $- 2$ by $1$ .

$0 - 2$

Step 2.2.9.1.4

Subtract $2$ from $0$ .

$- 2$

Step 2.2.9.2

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

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

Step 2.2.10

Simplify.

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Step 2.2.10.1

Move the negative in front of the fraction.

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

Step 2.2.10.2

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

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

Step 2.2.10.3

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

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

Step 2.2.11

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

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

Step 2.2.12

Multiply $- 1$ by $- 2$ .

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

Step 2.2.13

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

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

Step 2.2.14

Simplify.

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Step 2.2.14.1

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

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

Step 2.2.14.2

Cancel the common factor of $2$ .

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Step 2.2.14.2.1

Cancel the common factor.

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

Step 2.2.14.2.2

Rewrite the expression.

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

Step 2.2.14.3

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

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

Step 2.2.15

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

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

Step 2.2.16

Simplify.

$- \ln (| u_{1} |) + \ln (| u_{2} |) + C_{3} = \int d t$

Step 2.2.17

Substitute back in for each integration substitution variable.

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Step 2.2.17.1

Replace all occurrences of $u_{1}$ with $x - 1$ .

$- \ln (| x - 1 |) + \ln (| u_{2} |) + C_{3} = \int d t$

Step 2.2.17.2

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

$- \ln (| x - 1 |) + \ln (| 1 - 2 x |) + C_{3} = \int d t$

Step 2.2.18

Reorder terms.

$\ln (| 1 - 2 x |) - \ln (| x - 1 |) + C_{3} = \int d t$

Step 2.3

Apply the constant rule.

$\ln (| 1 - 2 x |) - \ln (| x - 1 |) + C_{3} = t + C_{4}$

Step 2.4

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

$\ln (| 1 - 2 x |) - \ln (| x - 1 |) = t + C$

Step 3

Solve for $x$ .

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Step 3.1

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

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

Step 3.2

Reorder $t$ and $C$ .

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

Step 3.3

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

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

Step 3.4

Rewrite $\ln (\frac{| 1 - 2 x |}{| x - 1 |}) = C + t$ 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 $b^{y} = x$ .

$e^{C + t} = \frac{| 1 - 2 x |}{| x - 1 |}$

Step 3.5

Solve for $x$ .

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Step 3.5.1

Rewrite the equation as $\frac{| 1 - 2 x |}{| x - 1 |} = e^{C + t}$ .

$\frac{| 1 - 2 x |}{| x - 1 |} = e^{C + t}$

Step 3.5.2

Multiply both sides by $| x - 1 |$ .

$\frac{| 1 - 2 x |}{| x - 1 |} | x - 1 | = e^{C + t} | x - 1 |$

Step 3.5.3

Simplify the left side.

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Step 3.5.3.1

Cancel the common factor of $| x - 1 |$ .

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Step 3.5.3.1.1

Cancel the common factor.

$\frac{| 1 - 2 x |}{| x - 1 |} | x - 1 | = e^{C + t} | x - 1 |$

Step 3.5.3.1.2

Rewrite the expression.

$| 1 - 2 x | = e^{C + t} | x - 1 |$

Step 3.5.4

Solve for $x$ .

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Step 3.5.4.1

Rewrite the equation as $e^{C + t} | x - 1 | = | 1 - 2 x |$ .

$e^{C + t} | x - 1 | = | 1 - 2 x |$

Step 3.5.4.2

Divide each term in $e^{C + t} | x - 1 | = | 1 - 2 x |$ by $e^{C + t}$ and simplify.

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Step 3.5.4.2.1

Divide each term in $e^{C + t} | x - 1 | = | 1 - 2 x |$ by $e^{C + t}$ .

$\frac{e^{C + t} | x - 1 |}{e^{C + t}} = \frac{| 1 - 2 x |}{e^{C + t}}$

Step 3.5.4.2.2

Simplify the left side.

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Step 3.5.4.2.2.1

Cancel the common factor of $e^{C + t}$ .

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Step 3.5.4.2.2.1.1

Cancel the common factor.

$\frac{e^{C + t} | x - 1 |}{e^{C + t}} = \frac{| 1 - 2 x |}{e^{C + t}}$

Step 3.5.4.2.2.1.2

Divide $| x - 1 |$ by $1$ .

$| x - 1 | = \frac{| 1 - 2 x |}{e^{C + t}}$

Step 3.5.4.3

Rewrite the absolute value equation as four equations without absolute value bars.

$x - 1 = \frac{1 - 2 x}{e^{C + t}}$

$x - 1 = - \frac{1 - 2 x}{e^{C + t}}$

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

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

Step 3.5.4.4

After simplifying, there are only two unique equations to be solved.

$x - 1 = \frac{1 - 2 x}{e^{C + t}}$

$x - 1 = - \frac{1 - 2 x}{e^{C + t}}$

Step 3.5.4.5

Solve $x - 1 = \frac{1 - 2 x}{e^{C + t}}$ for $x$ .

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Step 3.5.4.5.1

Multiply both sides by $e^{C + t}$ .

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

Step 3.5.4.5.2

Simplify.

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Step 3.5.4.5.2.1

Simplify the left side.

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Step 3.5.4.5.2.1.1

Simplify $(x - 1) e^{C + t}$ .

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Step 3.5.4.5.2.1.1.1

Apply the distributive property.

$x e^{C + t} - 1 e^{C + t} = \frac{1 - 2 x}{e^{C + t}} e^{C + t}$

Step 3.5.4.5.2.1.1.2

Rewrite $- 1 e^{C + t}$ as $- e^{C + t}$ .

$x e^{C + t} - e^{C + t} = \frac{1 - 2 x}{e^{C + t}} e^{C + t}$

Step 3.5.4.5.2.2

Simplify the right side.

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Step 3.5.4.5.2.2.1

Simplify $\frac{1 - 2 x}{e^{C + t}} e^{C + t}$ .

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Step 3.5.4.5.2.2.1.1

Cancel the common factor of $e^{C + t}$ .

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Step 3.5.4.5.2.2.1.1.1

Cancel the common factor.

$x e^{C + t} - e^{C + t} = \frac{1 - 2 x}{e^{C + t}} e^{C + t}$

Step 3.5.4.5.2.2.1.1.2

Rewrite the expression.

$x e^{C + t} - e^{C + t} = 1 - 2 x$

Step 3.5.4.5.2.2.1.2

Reorder $1$ and $- 2 x$ .

$x e^{C + t} - e^{C + t} = - 2 x + 1$

Step 3.5.4.5.3

Solve for $x$ .

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Step 3.5.4.5.3.1

Add $2 x$ to both sides of the equation.

$x e^{C + t} - e^{C + t} + 2 x = 1$

Step 3.5.4.5.3.2

Add $e^{C + t}$ to both sides of the equation.

$x e^{C + t} + 2 x = 1 + e^{C + t}$

Step 3.5.4.5.3.3

Factor $x$ out of $x e^{C + t} + 2 x$ .

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Step 3.5.4.5.3.3.1

Factor $x$ out of $x e^{C + t}$ .

$x e^{C + t} + 2 x = 1 + e^{C + t}$

Step 3.5.4.5.3.3.2

Factor $x$ out of $2 x$ .

$x e^{C + t} + x \cdot 2 = 1 + e^{C + t}$

Step 3.5.4.5.3.3.3

Factor $x$ out of $x e^{C + t} + x \cdot 2$ .

$x (e^{C + t} + 2) = 1 + e^{C + t}$

Step 3.5.4.5.3.4

Divide each term in $x (e^{C + t} + 2) = 1 + e^{C + t}$ by $e^{C + t} + 2$ and simplify.

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Step 3.5.4.5.3.4.1

Divide each term in $x (e^{C + t} + 2) = 1 + e^{C + t}$ by $e^{C + t} + 2$ .

$\frac{x (e^{C + t} + 2)}{e^{C + t} + 2} = \frac{1}{e^{C + t} + 2} + \frac{e^{C + t}}{e^{C + t} + 2}$

Step 3.5.4.5.3.4.2

Simplify the left side.

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Step 3.5.4.5.3.4.2.1

Cancel the common factor of $e^{C + t} + 2$ .

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Step 3.5.4.5.3.4.2.1.1

Cancel the common factor.

$\frac{x (e^{C + t} + 2)}{e^{C + t} + 2} = \frac{1}{e^{C + t} + 2} + \frac{e^{C + t}}{e^{C + t} + 2}$

Step 3.5.4.5.3.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{e^{C + t} + 2} + \frac{e^{C + t}}{e^{C + t} + 2}$

Step 3.5.4.5.3.4.3

Simplify the right side.

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Step 3.5.4.5.3.4.3.1

Combine the numerators over the common denominator.

$x = \frac{1 + e^{C + t}}{e^{C + t} + 2}$

Step 3.5.4.6

Solve $x - 1 = - \frac{1 - 2 x}{e^{C + t}}$ for $x$ .

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Step 3.5.4.6.1

Multiply both sides by $e^{C + t}$ .

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

Step 3.5.4.6.2

Simplify.

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Step 3.5.4.6.2.1

Simplify the left side.

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Step 3.5.4.6.2.1.1

Simplify $(x - 1) e^{C + t}$ .

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Step 3.5.4.6.2.1.1.1

Apply the distributive property.

$x e^{C + t} - 1 e^{C + t} = - \frac{1 - 2 x}{e^{C + t}} e^{C + t}$

Step 3.5.4.6.2.1.1.2

Rewrite $- 1 e^{C + t}$ as $- e^{C + t}$ .

$x e^{C + t} - e^{C + t} = - \frac{1 - 2 x}{e^{C + t}} e^{C + t}$

Step 3.5.4.6.2.2

Simplify the right side.

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Step 3.5.4.6.2.2.1

Simplify $- \frac{1 - 2 x}{e^{C + t}} e^{C + t}$ .

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Step 3.5.4.6.2.2.1.1

Cancel the common factor of $e^{C + t}$ .

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Step 3.5.4.6.2.2.1.1.1

Move the leading negative in $- \frac{1 - 2 x}{e^{C + t}}$ into the numerator.

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

Step 3.5.4.6.2.2.1.1.2

Cancel the common factor.

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

Step 3.5.4.6.2.2.1.1.3

Rewrite the expression.

$x e^{C + t} - e^{C + t} = - (1 - 2 x)$

Step 3.5.4.6.2.2.1.2

Apply the distributive property.

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

Step 3.5.4.6.2.2.1.3

Simplify the expression.

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Step 3.5.4.6.2.2.1.3.1

Multiply $- 1$ by $1$ .

$x e^{C + t} - e^{C + t} = - 1 - (- 2 x)$

Step 3.5.4.6.2.2.1.3.2

Multiply $- 2$ by $- 1$ .

$x e^{C + t} - e^{C + t} = - 1 + 2 x$

Step 3.5.4.6.2.2.1.3.3

Reorder $- 1$ and $2 x$ .

$x e^{C + t} - e^{C + t} = 2 x - 1$

Step 3.5.4.6.3

Solve for $x$ .

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Step 3.5.4.6.3.1

Subtract $2 x$ from both sides of the equation.

$x e^{C + t} - e^{C + t} - 2 x = - 1$

Step 3.5.4.6.3.2

Add $e^{C + t}$ to both sides of the equation.

$x e^{C + t} - 2 x = - 1 + e^{C + t}$

Step 3.5.4.6.3.3

Factor $x$ out of $x e^{C + t} - 2 x$ .

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Step 3.5.4.6.3.3.1

Factor $x$ out of $x e^{C + t}$ .

$x e^{C + t} - 2 x = - 1 + e^{C + t}$

Step 3.5.4.6.3.3.2

Factor $x$ out of $- 2 x$ .

$x e^{C + t} + x \cdot - 2 = - 1 + e^{C + t}$

Step 3.5.4.6.3.3.3

Factor $x$ out of $x e^{C + t} + x \cdot - 2$ .

$x (e^{C + t} - 2) = - 1 + e^{C + t}$

Step 3.5.4.6.3.4

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

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Step 3.5.4.6.3.4.1

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

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

Step 3.5.4.6.3.4.2

Simplify the left side.

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Step 3.5.4.6.3.4.2.1

Cancel the common factor of $e^{C + t} - 2$ .

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Step 3.5.4.6.3.4.2.1.1

Cancel the common factor.

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

Step 3.5.4.6.3.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{e^{C + t} - 2} + \frac{e^{C + t}}{e^{C + t} - 2}$

Step 3.5.4.6.3.4.3

Simplify the right side.

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Step 3.5.4.6.3.4.3.1

Combine the numerators over the common denominator.

$x = \frac{- 1 + e^{C + t}}{e^{C + t} - 2}$

Step 3.5.4.6.3.4.3.2

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

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

Step 3.5.4.6.3.4.3.3

Factor $- 1$ out of $e^{C + t}$ .

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

Step 3.5.4.6.3.4.3.4

Factor $- 1$ out of $- 1 (1) - 1 (- e^{C + t})$ .

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

Step 3.5.4.6.3.4.3.5

Move the negative in front of the fraction.

$x = - \frac{1 - e^{C + t}}{e^{C + t} - 2}$

Step 3.5.4.7

List all of the solutions.

$x = \frac{1 + e^{C + t}}{e^{C + t} + 2}$

$x = - \frac{1 - e^{C + t}}{e^{C + t} - 2}$

$x = \frac{1 + e^{C + t}}{e^{C + t} + 2}$

$x = - \frac{1 - e^{C + t}}{e^{C + t} - 2}$

$x = \frac{1 + e^{C + t}}{e^{C + t} + 2}$

$x = - \frac{1 - e^{C + t}}{e^{C + t} - 2}$

$x = \frac{1 + e^{C + t}}{e^{C + t} + 2}$

$x = - \frac{1 - e^{C + t}}{e^{C + t} - 2}$

Step 4

Group the constant terms together.

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Step 4.1

Reorder terms.

$x = \frac{1 + e^{t + C}}{e^{C + t} + 2}$

$x = - \frac{1 - e^{C + t}}{e^{C + t} - 2}$

Step 4.2

Rewrite $e^{t + C}$ as $e^{t} e^{C}$ .

$x = \frac{1 + e^{t} e^{C}}{e^{C + t} + 2}$

$x = - \frac{1 - e^{C + t}}{e^{C + t} - 2}$

Step 4.3

Reorder $e^{t}$ and $e^{C}$ .

$x = \frac{1 + e^{C} e^{t}}{e^{C + t} + 2}$

$x = - \frac{1 - e^{C + t}}{e^{C + t} - 2}$

Step 4.4

Reorder terms.

$x = \frac{1 + e^{C} e^{t}}{e^{t + C} + 2}$

$x = - \frac{1 - e^{C + t}}{e^{C + t} - 2}$

Step 4.5

Rewrite $e^{t + C}$ as $e^{t} e^{C}$ .

$x = \frac{1 + e^{C} e^{t}}{e^{t} e^{C} + 2}$

$x = - \frac{1 - e^{C + t}}{e^{C + t} - 2}$

Step 4.6

Reorder $e^{t}$ and $e^{C}$ .

$x = \frac{1 + e^{C} e^{t}}{e^{C} e^{t} + 2}$

$x = - \frac{1 - e^{C + t}}{e^{C + t} - 2}$

Step 4.7

Reorder terms.

$x = \frac{1 + e^{C} e^{t}}{e^{C} e^{t} + 2}$

$x = - \frac{1 - e^{t + C}}{e^{C + t} - 2}$

Step 4.8

Rewrite $e^{t + C}$ as $e^{t} e^{C}$ .

$x = \frac{1 + e^{C} e^{t}}{e^{C} e^{t} + 2}$

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

Step 4.9

Reorder $e^{t}$ and $e^{C}$ .

$x = \frac{1 + e^{C} e^{t}}{e^{C} e^{t} + 2}$

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

Step 4.10

Reorder terms.

$x = \frac{1 + e^{C} e^{t}}{e^{C} e^{t} + 2}$

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

Step 4.11

Rewrite $e^{t + C}$ as $e^{t} e^{C}$ .

$x = \frac{1 + e^{C} e^{t}}{e^{C} e^{t} + 2}$

$x = - \frac{1 - (e^{C} e^{t})}{e^{t} e^{C} - 2}$

Step 4.12

Reorder $e^{t}$ and $e^{C}$ .

$x = \frac{1 + e^{C} e^{t}}{e^{C} e^{t} + 2}$

$x = - \frac{1 - (e^{C} e^{t})}{e^{C} e^{t} - 2}$

$x = \frac{1 + e^{C} e^{t}}{e^{C} e^{t} + 2}$

$x = - \frac{1 - (e^{C} e^{t})}{e^{C} e^{t} - 2}$