Calculus Examples

$\frac{d y}{d x} = y + y^{3}$

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{y + y^{3}}$ .

$\frac{1}{y + y^{3}} \frac{d y}{d x} = \frac{1}{y + y^{3}} (y + y^{3})$

Step 1.2

Cancel the common factor of $y + y^{3}$ .

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Step 1.2.1

Cancel the common factor.

$\frac{1}{y + y^{3}} \frac{d y}{d x} = \frac{1}{y + y^{3}} (y + y^{3})$

Step 1.2.2

Rewrite the expression.

$\frac{1}{y + y^{3}} \frac{d y}{d x} = 1$

Step 1.3

Rewrite the equation.

$\frac{1}{y + y^{3}} d y = 1 d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y + y^{3}} d y = \int d x$

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

Factor $y$ out of $y + y^{3}$ .

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Step 2.2.1.1.1.1

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

$\frac{1}{y^{1} + y^{3}}$

Step 2.2.1.1.1.2

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

$\frac{1}{y \cdot 1 + y^{3}}$

Step 2.2.1.1.1.3

Factor $y$ out of $y^{3}$ .

$\frac{1}{y \cdot 1 + y \cdot y^{2}}$

Step 2.2.1.1.1.4

Factor $y$ out of $y \cdot 1 + y \cdot y^{2}$ .

$\frac{1}{y \cdot (1 + y^{2})}$

Step 2.2.1.1.1.5

Multiply $y$ by $1 + y^{2}$ .

$\frac{1}{y (1 + y^{2})}$

Step 2.2.1.1.2

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 is 2nd order, $2$ terms are required in the numerator. The number of terms required in the numerator is always equal to the order of the factor in the denominator.

$\frac{A}{y} + \frac{B y + C}{1 + y^{2}}$

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 $y (1 + y^{2})$ .

$\frac{1 (y (1 + y^{2}))}{y (1 + y^{2})} = \frac{A (y (1 + y^{2}))}{y} + \frac{(B y + C) (y (1 + y^{2}))}{1 + y^{2}}$

Step 2.2.1.1.4

Cancel the common factor of $y$ .

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Step 2.2.1.1.4.1

Cancel the common factor.

$\frac{1 (y (1 + y^{2}))}{y (1 + y^{2})} = \frac{A (y (1 + y^{2}))}{y} + \frac{(B y + C) (y (1 + y^{2}))}{1 + y^{2}}$

Step 2.2.1.1.4.2

Rewrite the expression.

$\frac{1 (1 + y^{2})}{1 + y^{2}} = \frac{A (y (1 + y^{2}))}{y} + \frac{(B y + C) (y (1 + y^{2}))}{1 + y^{2}}$

Step 2.2.1.1.5

Cancel the common factor of $1 + y^{2}$ .

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Step 2.2.1.1.5.1

Cancel the common factor.

$\frac{1 (1 + y^{2})}{1 + y^{2}} = \frac{A (y (1 + y^{2}))}{y} + \frac{(B y + C) (y (1 + y^{2}))}{1 + y^{2}}$

Step 2.2.1.1.5.2

Rewrite the expression.

$1 = \frac{A (y (1 + y^{2}))}{y} + \frac{(B y + C) (y (1 + y^{2}))}{1 + y^{2}}$

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 $y$ .

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Step 2.2.1.1.6.1.1

Cancel the common factor.

$1 = \frac{A (y (1 + y^{2}))}{y} + \frac{(B y + C) (y (1 + y^{2}))}{1 + y^{2}}$

Step 2.2.1.1.6.1.2

Divide $A (1 + y^{2})$ by $1$ .

$1 = A (1 + y^{2}) + \frac{(B y + C) (y (1 + y^{2}))}{1 + y^{2}}$

Step 2.2.1.1.6.2

Apply the distributive property.

$1 = A \cdot 1 + A y^{2} + \frac{(B y + C) (y (1 + y^{2}))}{1 + y^{2}}$

Step 2.2.1.1.6.3

Multiply $A$ by $1$ .

$1 = A + A y^{2} + \frac{(B y + C) (y (1 + y^{2}))}{1 + y^{2}}$

Step 2.2.1.1.6.4

Cancel the common factor of $1 + y^{2}$ .

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Step 2.2.1.1.6.4.1

Cancel the common factor.

$1 = A + A y^{2} + \frac{(B y + C) (y (1 + y^{2}))}{1 + y^{2}}$

Step 2.2.1.1.6.4.2

Divide $(B y + C) (y)$ by $1$ .

$1 = A + A y^{2} + (B y + C) (y)$

Step 2.2.1.1.6.5

Apply the distributive property.

$1 = A + A y^{2} + B y \cdot y + C y$

Step 2.2.1.1.6.6

Multiply $y$ by $y$ by adding the exponents.

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Step 2.2.1.1.6.6.1

Move $y$ .

$1 = A + A y^{2} + B (y \cdot y) + C y$

Step 2.2.1.1.6.6.2

Multiply $y$ by $y$ .

$1 = A + A y^{2} + B y^{2} + C y$

Step 2.2.1.1.7

Move $A$ .

$1 = A y^{2} + B y^{2} + C y + A$

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 $y^{2}$ 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 2.2.1.2.2

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

$0 = C$

Step 2.2.1.2.3

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

$1 = A$

Step 2.2.1.2.4

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

$0 = A + B$

$0 = C$

$1 = A$

$0 = A + B$

$0 = C$

$1 = A$

Step 2.2.1.3

Solve the system of equations.

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Step 2.2.1.3.1

Rewrite the equation as $C = 0$ .

$C = 0$

$0 = A + B$

$1 = A$

Step 2.2.1.3.2

Rewrite the equation as $A = 1$ .

$A = 1$

$C = 0$

$0 = A + B$

Step 2.2.1.3.3

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

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Step 2.2.1.3.3.1

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

$0 = (1) + B$

$A = 1$

$C = 0$

Step 2.2.1.3.3.2

Simplify the right side.

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Step 2.2.1.3.3.2.1

Remove parentheses.

$0 = 1 + B$

$A = 1$

$C = 0$

$0 = 1 + B$

$A = 1$

$C = 0$

$0 = 1 + B$

$A = 1$

$C = 0$

Step 2.2.1.3.4

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

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Step 2.2.1.3.4.1

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

$1 + B = 0$

$A = 1$

$C = 0$

Step 2.2.1.3.4.2

Subtract $1$ from both sides of the equation.

$B = - 1$

$A = 1$

$C = 0$

$B = - 1$

$A = 1$

$C = 0$

Step 2.2.1.3.5

Solve the system of equations.

$B = - 1 A = 1 C = 0$

Step 2.2.1.3.6

List all of the solutions.

$B = - 1, A = 1, C = 0$

Step 2.2.1.4

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

$\frac{1}{y} + \frac{- 1 y + 0}{1 + y^{2}}$

Step 2.2.1.5

Simplify.

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Step 2.2.1.5.1

Remove parentheses.

$\frac{1}{y} + \frac{(- 1) \cdot y + 0}{1 + y^{2}}$

Step 2.2.1.5.2

Simplify the numerator.

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Step 2.2.1.5.2.1

Rewrite $- 1 y$ as $- y$ .

$\frac{1}{y} + \frac{- y + 0}{1 + y^{2}}$

Step 2.2.1.5.2.2

Add $- y$ and $0$ .

$\frac{1}{y} + \frac{- y}{1 + y^{2}}$

Step 2.2.1.5.3

Move the negative in front of the fraction.

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

Step 2.2.2

Split the single integral into multiple integrals.

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Let $u = 1 + y^{2}$ . Then $d u = 2 y d y$ , so $\frac{1}{2} d u = y d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.5.1

Let $u = 1 + y^{2}$ . Find $\frac{d u}{d y}$ .

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Step 2.2.5.1.1

Differentiate $1 + y^{2}$ .

$\frac{d}{d y} [1 + y^{2}]$

Step 2.2.5.1.2

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

$\frac{d}{d y} [1] + \frac{d}{d y} [y^{2}]$

Step 2.2.5.1.3

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

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

Step 2.2.5.1.4

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

$0 + 2 y$

Step 2.2.5.1.5

Add $0$ and $2 y$ .

$2 y$

Step 2.2.5.2

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

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

Step 2.2.6

Simplify.

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Step 2.2.6.1

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

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

Step 2.2.6.2

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

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

Step 2.2.7

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

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

Step 2.2.8

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

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

Step 2.2.9

Simplify.

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

Step 2.2.10

Replace all occurrences of $u$ with $1 + y^{2}$ .

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

Step 2.2.11

Reorder terms.

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

Step 2.3

Apply the constant rule.

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

Step 2.4

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

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