Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

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

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

Step 1.2

Simplify.

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Step 1.2.1

Cancel the common factor of $y$ .

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Step 1.2.1.1

Cancel the common factor.

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

Step 1.2.1.2

Rewrite the expression.

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

Step 1.2.2

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

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Step 1.2.2.1

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

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{x \cdot x + x}$

Step 1.2.2.2

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

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{x \cdot x + x^{1}}$

Step 1.2.2.3

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

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{x \cdot x + x \cdot 1}$

Step 1.2.2.4

Factor $x$ out of $x \cdot x + x \cdot 1$ .

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Write the fraction using partial fraction decomposition.

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Step 2.3.1.1

Decompose the fraction and multiply through by the common denominator.

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Step 2.3.1.1.1

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 $B$ .

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

Step 2.3.1.1.2

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

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

Step 2.3.1.1.3

Cancel the common factor of $x$ .

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Step 2.3.1.1.3.1

Cancel the common factor.

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

Step 2.3.1.1.3.2

Rewrite the expression.

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

Step 2.3.1.1.4

Cancel the common factor of $x + 1$ .

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Step 2.3.1.1.4.1

Cancel the common factor.

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

Step 2.3.1.1.4.2

Rewrite the expression.

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

Step 2.3.1.1.5

Simplify each term.

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Step 2.3.1.1.5.1

Cancel the common factor of $x$ .

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Step 2.3.1.1.5.1.1

Cancel the common factor.

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

Step 2.3.1.1.5.1.2

Divide $A (x + 1)$ by $1$ .

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

Step 2.3.1.1.5.2

Apply the distributive property.

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

Step 2.3.1.1.5.3

Multiply $A$ by $1$ .

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

Step 2.3.1.1.5.4

Cancel the common factor of $x + 1$ .

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Step 2.3.1.1.5.4.1

Cancel the common factor.

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

Step 2.3.1.1.5.4.2

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

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

$1 = A x + A + B x$

Step 2.3.1.1.6

Move $A$ .

$1 = A x + B x + A$

Step 2.3.1.2

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

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Step 2.3.1.2.1

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.

$0 = A + B$

Step 2.3.1.2.2

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.

$1 = A$

Step 2.3.1.2.3

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

$0 = A + B$

$1 = A$

$0 = A + B$

$1 = A$

Step 2.3.1.3

Solve the system of equations.

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Step 2.3.1.3.1

Rewrite the equation as $A = 1$ .

$A = 1$

$0 = A + B$

Step 2.3.1.3.2

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

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Step 2.3.1.3.2.1

Replace all occurrences of $A$ in $0 = A + B$ with $1$ .

$0 = (1) + B$

$A = 1$

Step 2.3.1.3.2.2

Simplify the right side.

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Step 2.3.1.3.2.2.1

Remove parentheses.

$0 = 1 + B$

$A = 1$

$0 = 1 + B$

$A = 1$

$0 = 1 + B$

$A = 1$

Step 2.3.1.3.3

Solve for $B$ in $0 = 1 + B$ .

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Step 2.3.1.3.3.1

Rewrite the equation as $1 + B = 0$ .

$1 + B = 0$

$A = 1$

Step 2.3.1.3.3.2

Subtract $1$ from both sides of the equation.

$B = - 1$

$A = 1$

$B = - 1$

$A = 1$

Step 2.3.1.3.4

Solve the system of equations.

$B = - 1 A = 1$

Step 2.3.1.3.5

List all of the solutions.

$B = - 1, A = 1$

Step 2.3.1.4

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

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

Step 2.3.1.5

Move the negative in front of the fraction.

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

Step 2.3.2

Split the single integral into multiple integrals.

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

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

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Step 2.3.5.1

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

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Step 2.3.5.1.1

Differentiate $x + 1$ .

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

Step 2.3.5.1.2

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 2.3.5.1.3

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 2.3.5.1.4

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

$1 + 0$

Step 2.3.5.1.5

Add $1$ and $0$ .

$1$

Step 2.3.5.2

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

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

Step 2.3.6

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

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

Step 2.3.7

Simplify.

$\ln (| y |) + C_{1} = \ln (| x |) - \ln (| u |) + C_{4}$

Step 2.3.8

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

$\ln (| y |) + C_{1} = \ln (\frac{| x |}{| u |}) + C_{4}$

Step 2.3.9

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

$\ln (| y |) + C_{1} = \ln (\frac{| x |}{| x + 1 |}) + C_{4}$

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Multiply the numerator by the reciprocal of the denominator.

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

Step 3.4

Multiply $| y | \frac{| x + 1 |}{| x |}$ .

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Step 3.4.1

Combine $| y |$ and $\frac{| x + 1 |}{| x |}$ .

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

Step 3.4.2

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

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

Step 3.5

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

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

Step 3.6

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

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

Step 3.7

Solve for $y$ .

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Step 3.7.1

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

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

Step 3.7.2

Multiply both sides by $| x |$ .

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

Step 3.7.3

Simplify the left side.

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Step 3.7.3.1

Simplify $\frac{| y (x + 1) |}{| x |} | x |$ .

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Step 3.7.3.1.1

Cancel the common factor of $| x |$ .

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Step 3.7.3.1.1.1

Cancel the common factor.

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

Step 3.7.3.1.1.2

Rewrite the expression.

$| y (x + 1) | = e^{C} | x |$

Step 3.7.3.1.2

Apply the distributive property.

$| y x + y \cdot 1 | = e^{C} | x |$

Step 3.7.3.1.3

Multiply $y$ by $1$ .

$| y x + y | = e^{C} | x |$

Step 3.7.4

Solve for $y$ .

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Step 3.7.4.1

Reorder factors in $e^{C} | x |$ .

$| y x + y | = | x | e^{C}$

Step 3.7.4.2

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

$y x + y = \pm | x | e^{C}$

Step 3.7.4.3

Reorder factors in $\pm | x | e^{C}$ .

$y x + y = \pm e^{C} | x |$

Step 3.7.4.4

Factor $y$ out of $y x + y$ .

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Step 3.7.4.4.1

Factor $y$ out of $y x$ .

$y (x) + y = \pm e^{C} | x |$

Step 3.7.4.4.2

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

$y (x) + y = \pm e^{C} | x |$

Step 3.7.4.4.3

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

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

Step 3.7.4.4.4

Factor $y$ out of $y (x) + y \cdot 1$ .

$y (x + 1) = \pm e^{C} | x |$

Step 3.7.4.5

Divide each term in $y (x + 1) = \pm e^{C} | x |$ by $x + 1$ and simplify.

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Step 3.7.4.5.1

Divide each term in $y (x + 1) = \pm e^{C} | x |$ by $x + 1$ .

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

Step 3.7.4.5.2

Simplify the left side.

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Step 3.7.4.5.2.1

Cancel the common factor of $x + 1$ .

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Step 3.7.4.5.2.1.1

Cancel the common factor.

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

Step 3.7.4.5.2.1.2

Divide $y$ by $1$ .

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

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

$y = \frac{\pm C | x |}{x + 1}$

Step 4.2

Combine constants with the plus or minus.

$y = \frac{C | x |}{x + 1}$