Calculus Examples

$\frac{d y}{d x} = \frac{(x - 1) y^{5}}{x^{2} (2 y^{3} - y)}$

Step 1

Separate the variables.

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Step 1.1

Regroup factors.

$\frac{d y}{d x} = \frac{x - 1}{x^{2}} \cdot \frac{y^{5}}{2 y^{3} - y}$

Step 1.2

Multiply both sides by $\frac{2 y^{3} - y}{y^{5}}$ .

$\frac{2 y^{3} - y}{y^{5}} \frac{d y}{d x} = \frac{2 y^{3} - y}{y^{5}} (\frac{x - 1}{x^{2}} \cdot \frac{y^{5}}{2 y^{3} - y})$

Step 1.3

Simplify.

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Step 1.3.1

Factor $y$ out of $2 y^{3} - y$ .

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Step 1.3.1.1

Factor $y$ out of $2 y^{3}$ .

$\frac{2 y^{3} - y}{y^{5}} \frac{d y}{d x} = \frac{y (2 y^{2}) - y}{y^{5}} (\frac{x - 1}{x^{2}} \cdot \frac{y^{5}}{2 y^{3} - y})$

Step 1.3.1.2

Factor $y$ out of $- y$ .

$\frac{2 y^{3} - y}{y^{5}} \frac{d y}{d x} = \frac{y (2 y^{2}) + y \cdot - 1}{y^{5}} (\frac{x - 1}{x^{2}} \cdot \frac{y^{5}}{2 y^{3} - y})$

Step 1.3.1.3

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

$\frac{2 y^{3} - y}{y^{5}} \frac{d y}{d x} = \frac{y (2 y^{2} - 1)}{y^{5}} (\frac{x - 1}{x^{2}} \cdot \frac{y^{5}}{2 y^{3} - y})$

Step 1.3.2

Cancel the common factors.

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Step 1.3.2.1

Factor $y$ out of $y^{5}$ .

$\frac{2 y^{3} - y}{y^{5}} \frac{d y}{d x} = \frac{y (2 y^{2} - 1)}{y \cdot y^{4}} (\frac{x - 1}{x^{2}} \cdot \frac{y^{5}}{2 y^{3} - y})$

Step 1.3.2.2

Cancel the common factor.

$\frac{2 y^{3} - y}{y^{5}} \frac{d y}{d x} = \frac{y (2 y^{2} - 1)}{y \cdot y^{4}} (\frac{x - 1}{x^{2}} \cdot \frac{y^{5}}{2 y^{3} - y})$

Step 1.3.2.3

Rewrite the expression.

$\frac{2 y^{3} - y}{y^{5}} \frac{d y}{d x} = \frac{2 y^{2} - 1}{y^{4}} (\frac{x - 1}{x^{2}} \cdot \frac{y^{5}}{2 y^{3} - y})$

Step 1.3.3

Factor $y$ out of $2 y^{3} - y$ .

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Step 1.3.3.1

Factor $y$ out of $2 y^{3}$ .

$\frac{2 y^{3} - y}{y^{5}} \frac{d y}{d x} = \frac{2 y^{2} - 1}{y^{4}} (\frac{x - 1}{x^{2}} \cdot \frac{y^{5}}{y (2 y^{2}) - y})$

Step 1.3.3.2

Factor $y$ out of $- y$ .

$\frac{2 y^{3} - y}{y^{5}} \frac{d y}{d x} = \frac{2 y^{2} - 1}{y^{4}} (\frac{x - 1}{x^{2}} \cdot \frac{y^{5}}{y (2 y^{2}) + y \cdot - 1})$

Step 1.3.3.3

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

$\frac{2 y^{3} - y}{y^{5}} \frac{d y}{d x} = \frac{2 y^{2} - 1}{y^{4}} (\frac{x - 1}{x^{2}} \cdot \frac{y^{5}}{y (2 y^{2} - 1)})$

Step 1.3.4

Cancel the common factor of $y^{5}$ and $y$ .

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Step 1.3.4.1

Factor $y$ out of $y^{5}$ .

$\frac{2 y^{3} - y}{y^{5}} \frac{d y}{d x} = \frac{2 y^{2} - 1}{y^{4}} (\frac{x - 1}{x^{2}} \cdot \frac{y \cdot y^{4}}{y (2 y^{2} - 1)})$

Step 1.3.4.2

Cancel the common factors.

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Step 1.3.4.2.1

Cancel the common factor.

$\frac{2 y^{3} - y}{y^{5}} \frac{d y}{d x} = \frac{2 y^{2} - 1}{y^{4}} (\frac{x - 1}{x^{2}} \cdot \frac{y \cdot y^{4}}{y (2 y^{2} - 1)})$

Step 1.3.4.2.2

Rewrite the expression.

$\frac{2 y^{3} - y}{y^{5}} \frac{d y}{d x} = \frac{2 y^{2} - 1}{y^{4}} (\frac{x - 1}{x^{2}} \cdot \frac{y^{4}}{2 y^{2} - 1})$

Step 1.3.5

Multiply $\frac{x - 1}{x^{2}}$ by $\frac{y^{4}}{2 y^{2} - 1}$ .

$\frac{2 y^{3} - y}{y^{5}} \frac{d y}{d x} = \frac{2 y^{2} - 1}{y^{4}} \cdot \frac{(x - 1) y^{4}}{x^{2} (2 y^{2} - 1)}$

Step 1.3.6

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

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Step 1.3.6.1

Factor $2 y^{2} - 1$ out of $x^{2} (2 y^{2} - 1)$ .

$\frac{2 y^{3} - y}{y^{5}} \frac{d y}{d x} = \frac{2 y^{2} - 1}{y^{4}} \cdot \frac{(x - 1) y^{4}}{(2 y^{2} - 1) x^{2}}$

Step 1.3.6.2

Cancel the common factor.

$\frac{2 y^{3} - y}{y^{5}} \frac{d y}{d x} = \frac{2 y^{2} - 1}{y^{4}} \cdot \frac{(x - 1) y^{4}}{(2 y^{2} - 1) x^{2}}$

Step 1.3.6.3

Rewrite the expression.

$\frac{2 y^{3} - y}{y^{5}} \frac{d y}{d x} = \frac{1}{y^{4}} \cdot \frac{(x - 1) y^{4}}{x^{2}}$

Step 1.3.7

Cancel the common factor of $y^{4}$ .

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Step 1.3.7.1

Factor $y^{4}$ out of $(x - 1) y^{4}$ .

$\frac{2 y^{3} - y}{y^{5}} \frac{d y}{d x} = \frac{1}{y^{4}} \cdot \frac{y^{4} (x - 1)}{x^{2}}$

Step 1.3.7.2

Cancel the common factor.

$\frac{2 y^{3} - y}{y^{5}} \frac{d y}{d x} = \frac{1}{y^{4}} \cdot \frac{y^{4} (x - 1)}{x^{2}}$

Step 1.3.7.3

Rewrite the expression.

$\frac{2 y^{3} - y}{y^{5}} \frac{d y}{d x} = \frac{x - 1}{x^{2}}$

Step 1.4

Rewrite the equation.

$\frac{2 y^{3} - y}{y^{5}} d y = \frac{x - 1}{x^{2}} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{2 y^{3} - y}{y^{5}} d y = \int \frac{x - 1}{x^{2}} d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Simplify the expression.

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Step 2.2.1.1

Simplify.

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Step 2.2.1.1.1

Factor $y$ out of $2 y^{3} - y$ .

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Step 2.2.1.1.1.1

Factor $y$ out of $2 y^{3}$ .

$\int \frac{y (2 y^{2}) - y}{y^{5}} d y = \int \frac{x - 1}{x^{2}} d x$

Step 2.2.1.1.1.2

Factor $y$ out of $- y$ .

$\int \frac{y (2 y^{2}) + y \cdot - 1}{y^{5}} d y = \int \frac{x - 1}{x^{2}} d x$

Step 2.2.1.1.1.3

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

$\int \frac{y (2 y^{2} - 1)}{y^{5}} d y = \int \frac{x - 1}{x^{2}} d x$

Step 2.2.1.1.2

Cancel the common factors.

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Step 2.2.1.1.2.1

Factor $y$ out of $y^{5}$ .

$\int \frac{y (2 y^{2} - 1)}{y \cdot y^{4}} d y = \int \frac{x - 1}{x^{2}} d x$

Step 2.2.1.1.2.2

Cancel the common factor.

$\int \frac{y (2 y^{2} - 1)}{y \cdot y^{4}} d y = \int \frac{x - 1}{x^{2}} d x$

Step 2.2.1.1.2.3

Rewrite the expression.

$\int \frac{2 y^{2} - 1}{y^{4}} d y = \int \frac{x - 1}{x^{2}} d x$

Step 2.2.1.2

Apply basic rules of exponents.

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Step 2.2.1.2.1

Move $y^{4}$ out of the denominator by raising it to the $- 1$ power.

$\int (2 y^{2} - 1) {(y^{4})}^{- 1} d y = \int \frac{x - 1}{x^{2}} d x$

Step 2.2.1.2.2

Multiply the exponents in ${(y^{4})}^{- 1}$ .

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Step 2.2.1.2.2.1

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

$\int (2 y^{2} - 1) y^{4 \cdot - 1} d y = \int \frac{x - 1}{x^{2}} d x$

Step 2.2.1.2.2.2

Multiply $4$ by $- 1$ .

$\int (2 y^{2} - 1) y^{- 4} d y = \int \frac{x - 1}{x^{2}} d x$

Step 2.2.2

Multiply $(2 y^{2} - 1) y^{- 4}$ .

$\int 2 y^{2} y^{- 4} - 1 y^{- 4} d y = \int \frac{x - 1}{x^{2}} d x$

Step 2.2.3

Simplify.

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Step 2.2.3.1

Multiply $y^{2}$ by $y^{- 4}$ by adding the exponents.

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Step 2.2.3.1.1

Move $y^{- 4}$ .

$\int 2 (y^{- 4} y^{2}) - 1 y^{- 4} d y = \int \frac{x - 1}{x^{2}} d x$

Step 2.2.3.1.2

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

$\int 2 y^{- 4 + 2} - 1 y^{- 4} d y = \int \frac{x - 1}{x^{2}} d x$

Step 2.2.3.1.3

Add $- 4$ and $2$ .

$\int 2 y^{- 2} - 1 y^{- 4} d y = \int \frac{x - 1}{x^{2}} d x$

Step 2.2.3.2

Rewrite $- 1 y^{- 4}$ as $- y^{- 4}$ .

$\int 2 y^{- 2} - y^{- 4} d y = \int \frac{x - 1}{x^{2}} d x$

Step 2.2.4

Split the single integral into multiple integrals.

$\int 2 y^{- 2} d y + \int - y^{- 4} d y = \int \frac{x - 1}{x^{2}} d x$

Step 2.2.5

Since $2$ is constant with respect to $y$ , move $2$ out of the integral.

$2 \int y^{- 2} d y + \int - y^{- 4} d y = \int \frac{x - 1}{x^{2}} d x$

Step 2.2.6

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

$2 (- y^{- 1} + C_{1}) + \int - y^{- 4} d y = \int \frac{x - 1}{x^{2}} d x$

Step 2.2.7

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

$2 (- y^{- 1} + C_{1}) - \int y^{- 4} d y = \int \frac{x - 1}{x^{2}} d x$

Step 2.2.8

By the Power Rule, the integral of $y^{- 4}$ with respect to $y$ is $- \frac{1}{3} y^{- 3}$ .

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

Step 2.2.9

Simplify.

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Step 2.2.9.1

Simplify.

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Step 2.2.9.1.1

Combine $y^{- 3}$ and $\frac{1}{3}$ .

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

Step 2.2.9.1.2

Move $y^{- 3}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 2.2.9.2

Simplify.

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

Step 2.2.9.3

Simplify.

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Step 2.2.9.3.1

Multiply $- 1$ by $2$ .

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

Step 2.2.9.3.2

Combine $- 2$ and $\frac{1}{y}$ .

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

Step 2.2.9.3.3

Move the negative in front of the fraction.

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

Step 2.2.9.3.4

Multiply $- 1$ by $- 1$ .

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

Step 2.2.9.3.5

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Apply basic rules of exponents.

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Step 2.3.1.1

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

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

Step 2.3.1.2

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

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Step 2.3.1.2.1

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

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

Step 2.3.1.2.2

Multiply $2$ by $- 1$ .

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

Step 2.3.2

Multiply $(x - 1) x^{- 2}$ .

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

Step 2.3.3

Simplify.

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Step 2.3.3.1

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

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Step 2.3.3.1.1

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

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Step 2.3.3.1.1.1

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

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

Step 2.3.3.1.1.2

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

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

Step 2.3.3.1.2

Subtract $2$ from $1$ .

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

Step 2.3.3.2

Rewrite $- 1 x^{- 2}$ as $- x^{- 2}$ .

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

Step 2.3.4

Split the single integral into multiple integrals.

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

Step 2.3.5

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

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

Step 2.3.6

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

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

Step 2.3.7

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

$- \frac{2}{y} + \frac{1}{3 y^{3}} + C_{3} = \ln (| x |) + C_{4} - (- x^{- 1} + C_{5})$

Step 2.3.8

Simplify.

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Step 2.3.8.1

Simplify.

$- \frac{2}{y} + \frac{1}{3 y^{3}} + C_{3} = \ln (| x |) - - \frac{1}{x} + C_{6}$

Step 2.3.8.2

Simplify.

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Step 2.3.8.2.1

Multiply $- 1$ by $- 1$ .

$- \frac{2}{y} + \frac{1}{3 y^{3}} + C_{3} = \ln (| x |) + 1 \frac{1}{x} + C_{6}$

Step 2.3.8.2.2

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

$- \frac{2}{y} + \frac{1}{3 y^{3}} + C_{3} = \ln (| x |) + \frac{1}{x} + C_{6}$

Step 2.3.9

Reorder terms.

$- \frac{2}{y} + \frac{1}{3 y^{3}} + C_{3} = \frac{1}{x} + \ln (| x |) + C_{6}$

Step 2.4

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

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