Calculus Examples

$(1 + x^{3}) d y - x^{2} (y d) x = 0$

Step 1

Add $x^{2} y d x$ to both sides of the equation.

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

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

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Step 3.1.1

Factor $1 + x^{3}$ out of $(1 + x^{3}) y$ .

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

Step 3.1.2

Cancel the common factor.

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

Step 3.1.3

Rewrite the expression.

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

Step 3.2

Cancel the common factor of $y$ .

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Step 3.2.1

Factor $y$ out of $(1 + x^{3}) y$ .

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

Step 3.2.2

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

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

Step 3.2.3

Cancel the common factor.

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

Step 3.2.4

Rewrite the expression.

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

Step 3.3

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

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

Step 3.4

Simplify the denominator.

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Step 3.4.1

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

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

Step 3.4.2

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 $b = x$ .

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

Step 3.4.3

Simplify.

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Step 3.4.3.1

One to any power is one.

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

Step 3.4.3.2

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

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

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

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

Step 4.3

Integrate the right side.

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Step 4.3.1

Let $u = (1 + x) (1 - x + x^{2})$ . Then $d u = 3 x^{2} d x$ , so $\frac{1}{3} d u = x^{2} d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.3.1.1

Let $u = (1 + x) (1 - x + x^{2})$ . Find $\frac{d u}{d x}$ .

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Step 4.3.1.1.1

Differentiate $(1 + x) (1 - x + x^{2})$ .

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

Step 4.3.1.1.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = 1 + x$ and $g (x) = 1 - x + x^{2}$ .

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

Step 4.3.1.1.3

Differentiate.

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Step 4.3.1.1.3.1

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

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

Step 4.3.1.1.3.2

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

$(1 + x) (0 + \frac{d}{d x} [- x] + \frac{d}{d x} [x^{2}]) + (1 - x + x^{2}) \frac{d}{d x} [1 + x]$

Step 4.3.1.1.3.3

Add $0$ and $\frac{d}{d x} [- x]$ .

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

Step 4.3.1.1.3.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

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

Step 4.3.1.1.3.5

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

$(1 + x) (- 1 \cdot 1 + \frac{d}{d x} [x^{2}]) + (1 - x + x^{2}) \frac{d}{d x} [1 + x]$

Step 4.3.1.1.3.6

Multiply $- 1$ by $1$ .

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

Step 4.3.1.1.3.7

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

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

Step 4.3.1.1.3.8

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

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

Step 4.3.1.1.3.9

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

$(1 + x) (- 1 + 2 x) + (1 - x + x^{2}) (0 + \frac{d}{d x} [x])$

Step 4.3.1.1.3.10

Add $0$ and $\frac{d}{d x} [x]$ .

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

Step 4.3.1.1.3.11

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

$(1 + x) (- 1 + 2 x) + (1 - x + x^{2}) \cdot 1$

Step 4.3.1.1.3.12

Multiply $1 - x + x^{2}$ by $1$ .

$(1 + x) (- 1 + 2 x) + 1 - x + x^{2}$

Step 4.3.1.1.4

Simplify.

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Step 4.3.1.1.4.1

Apply the distributive property.

$(1 + x) \cdot - 1 + (1 + x) (2 x) + 1 - x + x^{2}$

Step 4.3.1.1.4.2

Apply the distributive property.

$1 \cdot - 1 + x \cdot - 1 + (1 + x) (2 x) + 1 - x + x^{2}$

Step 4.3.1.1.4.3

Apply the distributive property.

$1 \cdot - 1 + x \cdot - 1 + 1 (2 x) + x (2 x) + 1 - x + x^{2}$

Step 4.3.1.1.4.4

Combine terms.

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Step 4.3.1.1.4.4.1

Multiply $- 1$ by $1$ .

$- 1 + x \cdot - 1 + 1 (2 x) + x (2 x) + 1 - x + x^{2}$

Step 4.3.1.1.4.4.2

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

$- 1 - 1 \cdot x + 1 (2 x) + x (2 x) + 1 - x + x^{2}$

Step 4.3.1.1.4.4.3

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

$- 1 - x + 1 (2 x) + x (2 x) + 1 - x + x^{2}$

Step 4.3.1.1.4.4.4

Multiply $2$ by $1$ .

$- 1 - x + 2 x + x (2 x) + 1 - x + x^{2}$

Step 4.3.1.1.4.4.5

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

$- 1 - x + 2 x + 2 (x^{1} x) + 1 - x + x^{2}$

Step 4.3.1.1.4.4.6

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

$- 1 - x + 2 x + 2 (x^{1} x^{1}) + 1 - x + x^{2}$

Step 4.3.1.1.4.4.7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- 1 - x + 2 x + 2 x^{1 + 1} + 1 - x + x^{2}$

Step 4.3.1.1.4.4.8

Add $1$ and $1$ .

$- 1 - x + 2 x + 2 x^{2} + 1 - x + x^{2}$

Step 4.3.1.1.4.4.9

Add $- x$ and $2 x$ .

$- 1 + x + 2 x^{2} + 1 - x + x^{2}$

Step 4.3.1.1.4.4.10

Add $- 1$ and $1$ .

$x + 2 x^{2} + 0 - x + x^{2}$

Step 4.3.1.1.4.4.11

Add $x + 2 x^{2}$ and $0$ .

$x + 2 x^{2} - x + x^{2}$

Step 4.3.1.1.4.4.12

Subtract $x$ from $x$ .

$2 x^{2} + 0 + x^{2}$

Step 4.3.1.1.4.4.13

Add $2 x^{2}$ and $0$ .

$2 x^{2} + x^{2}$

Step 4.3.1.1.4.4.14

Add $2 x^{2}$ and $x^{2}$ .

$3 x^{2}$

Step 4.3.1.2

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

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

Step 4.3.2

Simplify.

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Step 4.3.2.1

Multiply $\frac{1}{u}$ by $\frac{1}{3}$ .

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

Step 4.3.2.2

Move $3$ to the left of $u$ .

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

Step 4.3.3

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

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

Step 4.3.4

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

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

Step 4.3.5

Simplify.

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

Step 4.3.6

Replace all occurrences of $u$ with $(1 + x) (1 - x + x^{2})$ .

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

Step 4.4

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

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

Step 5

Solve for $y$ .

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Step 5.1

Simplify the right side.

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Step 5.1.1

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

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

Step 5.2

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

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

Step 5.3

Simplify the numerator.

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Step 5.3.1

Expand $(1 + x) (1 - x + x^{2})$ by multiplying each term in the first expression by each term in the second expression.

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

Step 5.3.2

Simplify each term.

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Step 5.3.2.1

Multiply $1$ by $1$ .

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

Step 5.3.2.2

Multiply $- x$ by $1$ .

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

Step 5.3.2.3

Multiply $x^{2}$ by $1$ .

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

Step 5.3.2.4

Multiply $x$ by $1$ .

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

Step 5.3.2.5

Rewrite using the commutative property of multiplication.

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

Step 5.3.2.6

Multiply $x$ by $x$ by adding the exponents.

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Step 5.3.2.6.1

Move $x$ .

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

Step 5.3.2.6.2

Multiply $x$ by $x$ .

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

Step 5.3.2.7

Multiply $x$ by $x^{2}$ by adding the exponents.

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Step 5.3.2.7.1

Multiply $x$ by $x^{2}$ .

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Step 5.3.2.7.1.1

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

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

Step 5.3.2.7.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 5.3.2.7.2

Add $1$ and $2$ .

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

Step 5.3.3

Combine the opposite terms in $1 - x + x^{2} + x - x^{2} + x^{3}$ .

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Step 5.3.3.1

Add $- x$ and $x$ .

$\ln (| y |) - \frac{\ln (| 1 + x^{2} + 0 - x^{2} + x^{3} |)}{3} = C$

Step 5.3.3.2

Add $1 + x^{2}$ and $0$ .

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

Step 5.3.3.3

Subtract $x^{2}$ from $x^{2}$ .

$\ln (| y |) - \frac{\ln (| 1 + 0 + x^{3} |)}{3} = C$

Step 5.3.3.4

Add $1$ and $0$ .

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

Step 5.4

To write $\ln (| y |)$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

Step 5.5

Simplify terms.

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Step 5.5.1

Combine $\ln (| y |)$ and $\frac{3}{3}$ .

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

Step 5.5.2

Combine the numerators over the common denominator.

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

Step 5.6

Move $3$ to the left of $\ln (| y |)$ .

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

Step 5.7

Simplify the left side.

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Step 5.7.1

Simplify $\frac{3 \ln (| y |) - \ln (| 1 + x^{3} |)}{3}$ .

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Step 5.7.1.1

Simplify the numerator.

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Step 5.7.1.1.1

Simplify $3 \ln (| y |)$ by moving $3$ inside the logarithm.

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

Step 5.7.1.1.2

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

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

Step 5.7.1.1.3

Simplify the denominator.

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Step 5.7.1.1.3.1

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

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

Step 5.7.1.1.3.2

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 $b = x$ .

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

Step 5.7.1.1.3.3

Simplify.

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Step 5.7.1.1.3.3.1

One to any power is one.

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

Step 5.7.1.1.3.3.2

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

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

Step 5.7.1.1.3.4

Expand $(1 + x) (1 - x + x^{2})$ by multiplying each term in the first expression by each term in the second expression.

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

Step 5.7.1.1.3.5

Simplify each term.

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Step 5.7.1.1.3.5.1

Multiply $1$ by $1$ .

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

Step 5.7.1.1.3.5.2

Multiply $- x$ by $1$ .

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

Step 5.7.1.1.3.5.3

Multiply $x^{2}$ by $1$ .

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

Step 5.7.1.1.3.5.4

Multiply $x$ by $1$ .

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

Step 5.7.1.1.3.5.5

Rewrite using the commutative property of multiplication.

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

Step 5.7.1.1.3.5.6

Multiply $x$ by $x$ by adding the exponents.

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Step 5.7.1.1.3.5.6.1

Move $x$ .

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

Step 5.7.1.1.3.5.6.2

Multiply $x$ by $x$ .

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

Step 5.7.1.1.3.5.7

Multiply $x$ by $x^{2}$ by adding the exponents.

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Step 5.7.1.1.3.5.7.1

Multiply $x$ by $x^{2}$ .

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Step 5.7.1.1.3.5.7.1.1

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

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

Step 5.7.1.1.3.5.7.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 5.7.1.1.3.5.7.2

Add $1$ and $2$ .

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

Step 5.7.1.1.3.6

Combine the opposite terms in $1 - x + x^{2} + x - x^{2} + x^{3}$ .

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Step 5.7.1.1.3.6.1

Add $- x$ and $x$ .

$\frac{\ln (\frac{{| y |}^{3}}{| 1 + x^{2} + 0 - x^{2} + x^{3} |})}{3} = C$

Step 5.7.1.1.3.6.2

Add $1 + x^{2}$ and $0$ .

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

Step 5.7.1.1.3.6.3

Subtract $x^{2}$ from $x^{2}$ .

$\frac{\ln (\frac{{| y |}^{3}}{| 1 + 0 + x^{3} |})}{3} = C$

Step 5.7.1.1.3.6.4

Add $1$ and $0$ .

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

Step 5.7.1.1.3.7

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

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

Step 5.7.1.1.3.8

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 $b = x$ .

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

Step 5.7.1.1.3.9

Simplify.

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Step 5.7.1.1.3.9.1

One to any power is one.

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

Step 5.7.1.1.3.9.2

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

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

Step 5.7.1.2

Rewrite $\frac{\ln (\frac{{| y |}^{3}}{| (1 + x) (1 - x + x^{2}) |})}{3}$ as $\frac{1}{3} \ln (\frac{{| y |}^{3}}{| (1 + x) (1 - x + x^{2}) |})$ .

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

Step 5.7.1.3

Simplify $\frac{1}{3} \ln (\frac{{| y |}^{3}}{| (1 + x) (1 - x + x^{2}) |})$ by moving $\frac{1}{3}$ inside the logarithm.

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

Step 5.7.1.4

Apply the product rule to $\frac{{| y |}^{3}}{| (1 + x) (1 - x + x^{2}) |}$ .

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

Step 5.7.1.5

Simplify the numerator.

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Step 5.7.1.5.1

Multiply the exponents in ${({| y |}^{3})}^{\frac{1}{3}}$ .

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Step 5.7.1.5.1.1

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

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

Step 5.7.1.5.1.2

Cancel the common factor of $3$ .

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Step 5.7.1.5.1.2.1

Cancel the common factor.

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

Step 5.7.1.5.1.2.2

Rewrite the expression.

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

Step 5.7.1.5.2

Simplify.

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

Step 5.7.1.6

Simplify the denominator.

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Step 5.7.1.6.1

Expand $(1 + x) (1 - x + x^{2})$ by multiplying each term in the first expression by each term in the second expression.

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

Step 5.7.1.6.2

Simplify each term.

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Step 5.7.1.6.2.1

Multiply $1$ by $1$ .

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

Step 5.7.1.6.2.2

Multiply $- x$ by $1$ .

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

Step 5.7.1.6.2.3

Multiply $x^{2}$ by $1$ .

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

Step 5.7.1.6.2.4

Multiply $x$ by $1$ .

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

Step 5.7.1.6.2.5

Rewrite using the commutative property of multiplication.

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

Step 5.7.1.6.2.6

Multiply $x$ by $x$ by adding the exponents.

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Step 5.7.1.6.2.6.1

Move $x$ .

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

Step 5.7.1.6.2.6.2

Multiply $x$ by $x$ .

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

Step 5.7.1.6.2.7

Multiply $x$ by $x^{2}$ by adding the exponents.

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Step 5.7.1.6.2.7.1

Multiply $x$ by $x^{2}$ .

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Step 5.7.1.6.2.7.1.1

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

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

Step 5.7.1.6.2.7.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 5.7.1.6.2.7.2

Add $1$ and $2$ .

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

Step 5.7.1.6.3

Combine the opposite terms in $1 - x + x^{2} + x - x^{2} + x^{3}$ .

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Step 5.7.1.6.3.1

Add $- x$ and $x$ .

$\ln (\frac{| y |}{{| 1 + x^{2} + 0 - x^{2} + x^{3} |}^{\frac{1}{3}}}) = C$

Step 5.7.1.6.3.2

Add $1 + x^{2}$ and $0$ .

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

Step 5.7.1.6.3.3

Subtract $x^{2}$ from $x^{2}$ .

$\ln (\frac{| y |}{{| 1 + 0 + x^{3} |}^{\frac{1}{3}}}) = C$

Step 5.7.1.6.3.4

Add $1$ and $0$ .

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

Step 5.8

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

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

Step 5.9

Rewrite $\ln (\frac{| y |}{{| 1 + x^{3} |}^{\frac{1}{3}}}) = C$ 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 $b^{y} = x$ .

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

Step 5.10

Solve for $y$ .

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Step 5.10.1

Rewrite the equation as $\frac{| y |}{{| 1 + x^{3} |}^{\frac{1}{3}}} = e^{C}$ .

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

Step 5.10.2

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

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

Step 5.10.3

Simplify the left side.

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Step 5.10.3.1

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

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Step 5.10.3.1.1

Cancel the common factor.

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

Step 5.10.3.1.2

Rewrite the expression.

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

Step 5.10.4

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

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

Step 6

Group the constant terms together.

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Step 6.1

Simplify the constant of integration.

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

Step 6.2

Combine constants with the plus or minus.

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