Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Divide each term in ${(y \ln (x))}^{- 1} \frac{d y}{d x} = {(\frac{x}{y + 1})}^{2}$ by ${(y \ln (x))}^{- 1}$ and simplify.

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Step 1.1.1

Divide each term in ${(y \ln (x))}^{- 1} \frac{d y}{d x} = {(\frac{x}{y + 1})}^{2}$ by ${(y \ln (x))}^{- 1}$ .

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

Step 1.1.2

Simplify the left side.

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Step 1.1.2.1

Cancel the common factor of ${(y \ln (x))}^{- 1}$ .

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Step 1.1.2.1.1

Cancel the common factor.

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

Step 1.1.2.1.2

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

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

Step 1.1.3

Simplify the right side.

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Step 1.1.3.1

Move ${(y \ln (x))}^{- 1}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

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

Step 1.1.3.2

Apply the product rule to $\frac{x}{y + 1}$ .

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

Step 1.1.3.3

Multiply $\frac{x^{2}}{{(y + 1)}^{2}} (y \ln (x))$ .

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Step 1.1.3.3.1

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

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

Step 1.1.3.3.2

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

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

Step 1.2

Regroup factors.

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

Step 1.3

Multiply both sides by $\frac{{(y + 1)}^{2}}{y}$ .

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

Step 1.4

Simplify.

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Step 1.4.1

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

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

Step 1.4.2

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

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

Step 1.4.3

Combine.

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

Step 1.4.4

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

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Step 1.4.4.1

Cancel the common factor.

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

Step 1.4.4.2

Rewrite the expression.

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

Step 1.4.5

Cancel the common factor of $y$ .

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Step 1.4.5.1

Cancel the common factor.

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

Step 1.4.5.2

Divide $x^{2} \ln (x)$ by $1$ .

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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

Step 2.2.2

Apply the distributive property.

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

Step 2.2.3

Apply the distributive property.

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

Step 2.2.4

Apply the distributive property.

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

Step 2.2.5

Reorder $y$ and $1$ .

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

Step 2.2.6

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

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

Step 2.2.7

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

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

Step 2.2.8

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

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

Step 2.2.9

Simplify the expression.

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Step 2.2.9.1

Add $1$ and $1$ .

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

Step 2.2.9.2

Multiply $y$ by $1$ .

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

Step 2.2.9.3

Multiply $y$ by $1$ .

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

Step 2.2.9.4

Multiply $1$ by $1$ .

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

Step 2.2.10

Add $y$ and $y$ .

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

Step 2.2.11

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

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Step 2.2.11.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$y$

$0$

$y^{2}$

$2 y$

$1$

Step 2.2.11.2

Divide the highest order term in the dividend $y^{2}$ by the highest order term in divisor $y$ .

					$y$
	$y$	+	$0$		$y^{2}$	+	$2 y$	+	$1$

Step 2.2.11.3

Multiply the new quotient term by the divisor.

				$y$
$y$	+	$0$		$y^{2}$	+	$2 y$	+	$1$
			+	$y^{2}$	+	$0$

Step 2.2.11.4

The expression needs to be subtracted from the dividend, so change all the signs in $y^{2} + 0$

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

Step 2.2.11.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

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

Step 2.2.11.6

Pull the next terms from the original dividend down into the current dividend.

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

Step 2.2.11.7

Divide the highest order term in the dividend $2 y$ by the highest order term in divisor $y$ .

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

Step 2.2.11.8

Multiply the new quotient term by the divisor.

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

Step 2.2.11.9

The expression needs to be subtracted from the dividend, so change all the signs in $2 y + 0$

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

Step 2.2.11.10

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

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

Step 2.2.11.11

The final answer is the quotient plus the remainder over the divisor.

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

Step 2.2.12

Split the single integral into multiple integrals.

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

Step 2.2.13

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

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

Step 2.2.14

Apply the constant rule.

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

Step 2.2.15

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

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

Step 2.2.16

Simplify.

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \ln (x)$ and $d v = x^{2}$ .

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

Step 2.3.2

Simplify.

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Step 2.3.2.1

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

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

Step 2.3.2.2

Combine $\ln (x)$ and $\frac{x^{3}}{3}$ .

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

Step 2.3.3

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

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

Step 2.3.4

Simplify.

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Step 2.3.4.1

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

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

Step 2.3.4.2

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

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Step 2.3.4.2.1

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

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

Step 2.3.4.2.2

Cancel the common factors.

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Step 2.3.4.2.2.1

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

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

Step 2.3.4.2.2.2

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

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

Step 2.3.4.2.2.3

Cancel the common factor.

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

Step 2.3.4.2.2.4

Rewrite the expression.

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

Step 2.3.4.2.2.5

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

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

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

Step 2.3.5

By the Power Rule, the integral of $x^{2}$ with respect to $x$ is $\frac{1}{3} x^{3}$ .

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

Step 2.3.6

Simplify the answer.

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Step 2.3.6.1

Rewrite $\frac{\ln (x) x^{3}}{3} - \frac{1}{3} (\frac{1}{3} x^{3} + C_{5})$ as $\frac{1}{3} \ln (x) x^{3} - \frac{1}{3} \cdot \frac{1}{3} x^{3} + C_{5}$ .

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

Step 2.3.6.2

Simplify.

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Step 2.3.6.2.1

Combine $\frac{1}{3}$ and $\ln (x)$ .

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

Step 2.3.6.2.2

Combine $\frac{\ln (x)}{3}$ and $x^{3}$ .

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

Step 2.3.6.2.3

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

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

Step 2.3.6.2.4

Multiply $3$ by $3$ .

$\frac{1}{2} y^{2} + 2 y + \ln (| y |) + C_{4} = \frac{\ln (x) x^{3}}{3} - \frac{1}{9} x^{3} + C_{5}$

Step 2.3.6.3

Combine $x^{3}$ and $\frac{1}{9}$ .

$\frac{1}{2} y^{2} + 2 y + \ln (| y |) + C_{4} = \frac{\ln (x) x^{3}}{3} - \frac{x^{3}}{9} + C_{5}$

Step 2.3.6.4

Reorder terms.

$\frac{1}{2} y^{2} + 2 y + \ln (| y |) + C_{4} = \frac{1}{3} \ln (x) x^{3} - \frac{1}{9} x^{3} + C_{5}$

Step 2.3.7

Reorder terms.

$\frac{1}{2} y^{2} + 2 y + \ln (| y |) + C_{4} = \frac{1}{3} x^{3} \ln (x) - \frac{1}{9} x^{3} + C_{5}$

Step 2.4

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

$\frac{1}{2} y^{2} + 2 y + \ln (| y |) = \frac{1}{3} x^{3} \ln (x) - \frac{1}{9} x^{3} + C$