Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{2 - y}$ .

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

Step 1.2

Cancel the common factor of $2 - y$ .

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Step 1.2.1

Cancel the common factor.

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

Step 1.2.2

Rewrite the expression.

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Rewrite.

$\frac{- 1}{1}$

Step 2.2.1.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 2.2.1.2

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

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

Step 2.2.2

Split the fraction into multiple fractions.

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Simplify.

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

Step 2.2.6

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Let $u_{2} = x + 1$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 2.3.1.1

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

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Step 2.3.1.1.1

Differentiate $x + 1$ .

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

Step 2.3.1.1.2

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 2.3.1.1.3

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 2.3.1.1.4

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

$1 + 0$

Step 2.3.1.1.5

Add $1$ and $0$ .

$1$

Step 2.3.1.2

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

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

Step 2.3.2

Apply basic rules of exponents.

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Step 2.3.2.1

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

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

Step 2.3.2.2

Multiply the exponents in ${({u_{2}}^{2})}^{- 1}$ .

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Step 2.3.2.2.1

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

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

Step 2.3.2.2.2

Multiply $2$ by $- 1$ .

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

Step 2.3.3

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

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

Step 2.3.4

Rewrite $- {u_{2}}^{- 1} + C_{2}$ as $- \frac{1}{u_{2}} + C_{2}$ .

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

Step 2.3.5

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Divide each term in $- \ln (| 2 - y |) = - \frac{1}{x + 1} + C$ by $- 1$ and simplify.

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Step 3.1.1

Divide each term in $- \ln (| 2 - y |) = - \frac{1}{x + 1} + C$ by $- 1$ .

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

Step 3.1.2

Simplify the left side.

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Step 3.1.2.1

Dividing two negative values results in a positive value.

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

Step 3.1.2.2

Divide $\ln (| 2 - y |)$ by $1$ .

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

Step 3.1.3

Simplify the right side.

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Step 3.1.3.1

Combine the numerators over the common denominator.

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

Step 3.1.3.2

To write $C$ as a fraction with a common denominator, multiply by $\frac{x + 1}{x + 1}$ .

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

Step 3.1.3.3

Combine the numerators over the common denominator.

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

Step 3.1.3.4

Simplify the numerator.

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Step 3.1.3.4.1

Apply the distributive property.

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

Step 3.1.3.4.2

Multiply $C$ by $1$ .

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

Step 3.1.3.5

Simplify terms.

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Step 3.1.3.5.1

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

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

Step 3.1.3.5.2

Factor $- 1$ out of $C x$ .

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

Step 3.1.3.5.3

Factor $- 1$ out of $- 1 (1) - (- C x)$ .

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

Step 3.1.3.5.4

Factor $- 1$ out of $C$ .

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

Step 3.1.3.5.5

Factor $- 1$ out of $- 1 (1 - C x) - 1 (- C)$ .

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

Step 3.1.3.5.6

Move the negative in front of the fraction.

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

Step 3.1.3.5.7

Dividing two negative values results in a positive value.

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

Step 3.1.3.5.8

Divide $\frac{1 - C x - C}{x + 1}$ by $1$ .

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

Solve for $y$ .

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Step 3.4.1

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

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

Step 3.4.2

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

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

Step 3.4.3

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

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Step 3.4.3.1

Subtract $2$ from both sides of the equation.

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

Step 3.4.3.2

Simplify each term.

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Step 3.4.3.2.1

Split the fraction $\frac{1 - C x - C}{x + 1}$ into two fractions.

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

Step 3.4.3.2.2

Simplify each term.

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Step 3.4.3.2.2.1

Split the fraction $\frac{1 - C x}{x + 1}$ into two fractions.

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

Step 3.4.3.2.2.2

Move the negative in front of the fraction.

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

Step 3.4.3.2.2.3

Move the negative in front of the fraction.

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

Step 3.4.4

Divide each term in $- y = \pm e^{\frac{1}{x + 1} - \frac{C x}{x + 1} - \frac{C}{x + 1}} - 2$ by $- 1$ and simplify.

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Step 3.4.4.1

Divide each term in $- y = \pm e^{\frac{1}{x + 1} - \frac{C x}{x + 1} - \frac{C}{x + 1}} - 2$ by $- 1$ .

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

Step 3.4.4.2

Simplify the left side.

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Step 3.4.4.2.1

Dividing two negative values results in a positive value.

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

Step 3.4.4.2.2

Divide $y$ by $1$ .

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

Step 3.4.4.3

Simplify the right side.

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Step 3.4.4.3.1

To write $\frac{\pm e^{\frac{1}{x + 1} - \frac{C x}{x + 1} - \frac{C}{x + 1}}}{- 1}$ as a fraction with a common denominator, multiply by $\frac{- 1}{- 1}$ .

$y = \frac{\pm e^{\frac{1}{x + 1} - \frac{C x}{x + 1} - \frac{C}{x + 1}}}{- 1} \cdot \frac{- 1}{- 1} + \frac{- 2}{- 1}$

Step 3.4.4.3.2

To write $\frac{- 2}{- 1}$ as a fraction with a common denominator, multiply by $\frac{- 1}{- 1}$ .

$y = \frac{\pm e^{\frac{1}{x + 1} - \frac{C x}{x + 1} - \frac{C}{x + 1}}}{- 1} \cdot \frac{- 1}{- 1} + \frac{- 2}{- 1} \cdot \frac{- 1}{- 1}$

Step 3.4.4.3.3

Write each expression with a common denominator of $1$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.4.4.3.3.1

Multiply $\frac{\pm e^{\frac{1}{x + 1} - \frac{C x}{x + 1} - \frac{C}{x + 1}}}{- 1}$ by $\frac{- 1}{- 1}$ .

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

Step 3.4.4.3.3.2

Multiply $- 1$ by $- 1$ .

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

Step 3.4.4.3.3.3

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

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

Step 3.4.4.3.3.4

Multiply $- 1$ by $- 1$ .

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

Step 3.4.4.3.4

Combine the numerators over the common denominator.

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

Step 3.4.4.3.5

Simplify each term.

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Step 3.4.4.3.5.1

Combine the numerators over the common denominator.

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

Step 3.4.4.3.5.2

Combine the numerators over the common denominator.

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

Step 3.4.4.3.5.3

Move $- 1$ to the left of $\pm e^{\frac{1 - C x - C}{x + 1}}$ .

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

Step 3.4.4.3.5.4

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

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

Step 3.4.4.3.5.5

Multiply $- 2$ by $- 1$ .

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

Step 3.4.4.3.6

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

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

Step 4

Simplify the constant of integration.

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