Calculus Examples

$\frac{d y}{d x} + \frac{3 x^{2}}{1 + x^{3}} y = \frac{1}{1 + x^{3}}$

Step 1

The integrating factor is defined by the formula $e^{\int P (x) d x}$ , where $P (x) = \frac{3 x^{2}}{1 + x^{3}}$ .

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Step 1.1

Set up the integration.

$e^{\int \frac{3 x^{2}}{1 + x^{3}} d x}$

Step 1.2

Integrate $\frac{3 x^{2}}{1 + x^{3}}$ .

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Step 1.2.1

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

$e^{3 \int \frac{x^{2}}{1 + x^{3}} d x}$

Step 1.2.2

Rewrite $x^{3}$ as ${(x^{3})}^{1}$ .

$e^{3 \int \frac{x^{2}}{1 + {(x^{3})}^{1}} d x}$

Step 1.2.3

Let $u_{1} = x^{3}$ . Then $d u_{1} = 3 x^{2} d x$ , so $\frac{1}{3} d u_{1} = x^{2} d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 1.2.3.1

Let $u_{1} = x^{3}$ . Find $\frac{d u_{1}}{d x}$ .

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Step 1.2.3.1.1

Differentiate $x^{3}$ .

$\frac{d}{d x} [x^{3}]$

Step 1.2.3.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$3 x^{2}$

Step 1.2.3.2

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

$e^{3 \int \frac{1}{1 + {u_{1}}^{1}} \cdot \frac{1}{3} d u_{1}}$

Step 1.2.4

Simplify.

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Step 1.2.4.1

Simplify.

$e^{3 \int \frac{1}{1 + u_{1}} \cdot \frac{1}{3} d u_{1}}$

Step 1.2.4.2

Multiply $\frac{1}{1 + u_{1}}$ by $\frac{1}{3}$ .

$e^{3 \int \frac{1}{(1 + u_{1}) \cdot 3} d u_{1}}$

Step 1.2.4.3

Move $3$ to the left of $1 + u_{1}$ .

$e^{3 \int \frac{1}{3 (1 + u_{1})} d u_{1}}$

Step 1.2.5

Since $\frac{1}{3}$ is constant with respect to $u_{1}$ , move $\frac{1}{3}$ out of the integral.

$e^{3 (\frac{1}{3} \int \frac{1}{1 + u_{1}} d u_{1})}$

Step 1.2.6

Simplify.

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Step 1.2.6.1

Combine $\frac{1}{3}$ and $3$ .

$e^{\frac{3}{3} \int \frac{1}{1 + u_{1}} d u_{1}}$

Step 1.2.6.2

Cancel the common factor of $3$ .

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Step 1.2.6.2.1

Cancel the common factor.

$e^{\frac{3}{3} \int \frac{1}{1 + u_{1}} d u_{1}}$

Step 1.2.6.2.2

Rewrite the expression.

$e^{1 \int \frac{1}{1 + u_{1}} d u_{1}}$

Step 1.2.6.3

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

$e^{\int \frac{1}{1 + u_{1}} d u_{1}}$

Step 1.2.7

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

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Step 1.2.7.1

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

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Step 1.2.7.1.1

Differentiate $1 + u_{1}$ .

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

Step 1.2.7.1.2

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

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

Step 1.2.7.1.3

Since $1$ is constant with respect to $u_{1}$ , the derivative of $1$ with respect to $u_{1}$ is $0$ .

$0 + \frac{d}{d u_{1}} [u_{1}]$

Step 1.2.7.1.4

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

$0 + 1$

Step 1.2.7.1.5

Add $0$ and $1$ .

$1$

Step 1.2.7.2

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

$e^{\int \frac{1}{u_{2}} d u_{2}}$

Step 1.2.8

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

$e^{\ln (| u_{2} |) + C}$

Step 1.2.9

Substitute back in for each integration substitution variable.

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Step 1.2.9.1

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

$e^{\ln (| 1 + u_{1} |) + C}$

Step 1.2.9.2

Replace all occurrences of $u_{1}$ with $x^{3}$ .

$e^{\ln (| 1 + x^{3} |) + C}$

Step 1.3

Remove the constant of integration.

$e^{\ln (1 + x^{3})}$

Step 1.4

Exponentiation and log are inverse functions.

$1 + x^{3}$

Step 2

Multiply each term by the integrating factor $1 + x^{3}$ .

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Step 2.1

Multiply each term by $1 + x^{3}$ .

$(1 + x^{3}) \frac{d y}{d x} + (1 + x^{3}) (\frac{3 x^{2}}{1 + x^{3}} y) = (1 + x^{3}) \frac{1}{1 + x^{3}}$

Step 2.2

Simplify each term.

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Step 2.2.1

Apply the distributive property.

$1 \frac{d y}{d x} + x^{3} \frac{d y}{d x} + (1 + x^{3}) (\frac{3 x^{2}}{1 + x^{3}} y) = (1 + x^{3}) \frac{1}{1 + x^{3}}$

Step 2.2.2

Multiply $\frac{d y}{d x}$ by $1$ .

$\frac{d y}{d x} + x^{3} \frac{d y}{d x} + (1 + x^{3}) (\frac{3 x^{2}}{1 + x^{3}} y) = (1 + x^{3}) \frac{1}{1 + x^{3}}$

Step 2.2.3

Simplify the denominator.

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Step 2.2.3.1

Rewrite $1$ as $1^{3}$ .

$\frac{d y}{d x} + x^{3} \frac{d y}{d x} + (1 + x^{3}) (\frac{3 x^{2}}{1^{3} + x^{3}} y) = (1 + x^{3}) \frac{1}{1 + x^{3}}$

Step 2.2.3.2

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = 1$ and $b = x$ .

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

Step 2.2.3.3

Simplify.

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Step 2.2.3.3.1

One to any power is one.

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

Step 2.2.3.3.2

Rewrite $- 1 x$ as $- x$ .

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

Step 2.2.4

Combine $\frac{3 x^{2}}{(1 + x) (1 - x + x^{2})}$ and $y$ .

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

Step 2.2.5

Multiply $1 + x^{3}$ by $\frac{3 x^{2} y}{(1 + x) (1 - x + x^{2})}$ .

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

Step 2.2.6

Simplify the numerator.

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Step 2.2.6.1

Rewrite $1$ as $1^{3}$ .

$\frac{d y}{d x} + x^{3} \frac{d y}{d x} + \frac{(1^{3} + x^{3}) \cdot 3 x^{2} y}{(1 + x) (1 - x + x^{2})} = (1 + x^{3}) \frac{1}{1 + x^{3}}$

Step 2.2.6.2

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = 1$ and $b = x$ .

$\frac{d y}{d x} + x^{3} \frac{d y}{d x} + \frac{(1 + x) (1^{2} - 1 x + x^{2}) \cdot 3 x^{2} y}{(1 + x) (1 - x + x^{2})} = (1 + x^{3}) \frac{1}{1 + x^{3}}$

Step 2.2.6.3

Simplify.

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Step 2.2.6.3.1

One to any power is one.

$\frac{d y}{d x} + x^{3} \frac{d y}{d x} + \frac{(1 + x) (1 - 1 x + x^{2}) \cdot 3 x^{2} y}{(1 + x) (1 - x + x^{2})} = (1 + x^{3}) \frac{1}{1 + x^{3}}$

Step 2.2.6.3.2

Rewrite $- 1 x$ as $- x$ .

$\frac{d y}{d x} + x^{3} \frac{d y}{d x} + \frac{(1 + x) (1 - x + x^{2}) \cdot 3 x^{2} y}{(1 + x) (1 - x + x^{2})} = (1 + x^{3}) \frac{1}{1 + x^{3}}$

Step 2.2.7

Cancel the common factor of $1 + x$ .

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Step 2.2.7.1

Cancel the common factor.

$\frac{d y}{d x} + x^{3} \frac{d y}{d x} + \frac{(1 + x) (1 - x + x^{2}) \cdot 3 x^{2} y}{(1 + x) (1 - x + x^{2})} = (1 + x^{3}) \frac{1}{1 + x^{3}}$

Step 2.2.7.2

Rewrite the expression.

$\frac{d y}{d x} + x^{3} \frac{d y}{d x} + \frac{(1 - x + x^{2}) \cdot 3 x^{2} y}{1 - x + x^{2}} = (1 + x^{3}) \frac{1}{1 + x^{3}}$

Step 2.2.8

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

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Step 2.2.8.1

Cancel the common factor.

$\frac{d y}{d x} + x^{3} \frac{d y}{d x} + \frac{(1 - x + x^{2}) \cdot 3 x^{2} y}{1 - x + x^{2}} = (1 + x^{3}) \frac{1}{1 + x^{3}}$

Step 2.2.8.2

Divide $3 x^{2} y$ by $1$ .

$\frac{d y}{d x} + x^{3} \frac{d y}{d x} + 3 x^{2} y = (1 + x^{3}) \frac{1}{1 + x^{3}}$

Step 2.3

Simplify the denominator.

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Step 2.3.1

Rewrite $1$ as $1^{3}$ .

$\frac{d y}{d x} + x^{3} \frac{d y}{d x} + 3 x^{2} y = (1 + x^{3}) \frac{1}{1^{3} + x^{3}}$

Step 2.3.2

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = 1$ and $b = x$ .

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

Step 2.3.3

Simplify.

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Step 2.3.3.1

One to any power is one.

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

Step 2.3.3.2

Rewrite $- 1 x$ as $- x$ .

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

Step 2.4

Multiply $1 + x^{3}$ by $\frac{1}{(1 + x) (1 - x + x^{2})}$ .

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

Step 2.5

Simplify the numerator.

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Step 2.5.1

Rewrite $1$ as $1^{3}$ .

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

Step 2.5.2

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = 1$ and $b = x$ .

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

Step 2.5.3

Simplify.

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Step 2.5.3.1

One to any power is one.

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

Step 2.5.3.2

Rewrite $- 1 x$ as $- x$ .

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

Step 2.6

Cancel the common factor of $1 + x$ .

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Step 2.6.1

Cancel the common factor.

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

Step 2.6.2

Rewrite the expression.

$\frac{d y}{d x} + x^{3} \frac{d y}{d x} + 3 x^{2} y = \frac{1 - x + x^{2}}{1 - x + x^{2}}$

Step 2.7

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

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Step 2.7.1

Cancel the common factor.

$\frac{d y}{d x} + x^{3} \frac{d y}{d x} + 3 x^{2} y = \frac{1 - x + x^{2}}{1 - x + x^{2}}$

Step 2.7.2

Rewrite the expression.

$\frac{d y}{d x} + x^{3} \frac{d y}{d x} + 3 x^{2} y = 1$

Step 3

Rewrite the left side as a result of differentiating a product.

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

Step 4

Set up an integral on each side.

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

Step 5

Integrate the left side.

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

Step 6

Apply the constant rule.

$(1 + x^{3}) y = x + C$

Step 7

Divide each term in $(1 + x^{3}) y = x + C$ by $1 + x^{3}$ and simplify.

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Step 7.1

Divide each term in $(1 + x^{3}) y = x + C$ by $1 + x^{3}$ .

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

Step 7.2

Simplify the left side.

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Step 7.2.1

Cancel the common factor of $1 + x^{3}$ .

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Step 7.2.1.1

Cancel the common factor.

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

Step 7.2.1.2

Divide $y$ by $1$ .

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

Step 7.3

Simplify the right side.

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Step 7.3.1

Simplify each term.

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Step 7.3.1.1

Simplify the denominator.

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Step 7.3.1.1.1

Rewrite $1$ as $1^{3}$ .

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

Step 7.3.1.1.2

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = 1$ and $b = x$ .

$y = \frac{x}{(1 + x) (1^{2} - 1 x + x^{2})} + \frac{C}{1 + x^{3}}$

Step 7.3.1.1.3

Simplify.

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Step 7.3.1.1.3.1

One to any power is one.

$y = \frac{x}{(1 + x) (1 - 1 x + x^{2})} + \frac{C}{1 + x^{3}}$

Step 7.3.1.1.3.2

Rewrite $- 1 x$ as $- x$ .

$y = \frac{x}{(1 + x) (1 - x + x^{2})} + \frac{C}{1 + x^{3}}$

Step 7.3.1.2

Simplify the denominator.

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Step 7.3.1.2.1

Rewrite $1$ as $1^{3}$ .

$y = \frac{x}{(1 + x) (1 - x + x^{2})} + \frac{C}{1^{3} + x^{3}}$

Step 7.3.1.2.2

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = 1$ and $b = x$ .

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

Step 7.3.1.2.3

Simplify.

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Step 7.3.1.2.3.1

One to any power is one.

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

Step 7.3.1.2.3.2

Rewrite $- 1 x$ as $- x$ .

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