Calculus Examples

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

Step 1

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

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $x + 1$ .

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Step 3.1.1

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

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

Step 3.1.2

Cancel the common factor.

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

Step 3.1.3

Rewrite the expression.

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

Step 3.2

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

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

Cancel the common factor.

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

Step 3.3.4

Rewrite the expression.

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

Step 3.4

Combine $\frac{1}{x + 1}$ and $x$ .

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

Apply basic rules of exponents.

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Step 4.2.1.1

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

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

Step 4.2.1.2

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

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Step 4.2.1.2.1

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

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

Step 4.2.1.2.2

Multiply $2$ by $- 1$ .

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

Step 4.2.2

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

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

Step 4.2.3

Simplify.

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Step 4.2.3.1

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

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Step 4.2.3.1.1

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

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Step 4.2.3.1.1.1

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

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

Step 4.2.3.1.1.2

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

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

Step 4.2.3.1.2

Subtract $2$ from $1$ .

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

Step 4.2.3.2

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

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

Step 4.2.4

Split the single integral into multiple integrals.

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

Step 4.2.5

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

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

Step 4.2.6

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

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

Step 4.2.7

Simplify.

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

Step 4.2.8

Reorder terms.

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

Step 4.3

Integrate the right side.

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Step 4.3.1

Divide $x$ by $x + 1$ .

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Step 4.3.1.1

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

$x$

$1$

$x$

$0$

Step 4.3.1.2

Divide the highest order term in the dividend $x$ by the highest order term in divisor $x$ .

					$1$
	$x$	+	$1$		$x$	+	$0$

Step 4.3.1.3

Multiply the new quotient term by the divisor.

				$1$
$x$	+	$1$		$x$	+	$0$
			+	$x$	+	$1$

Step 4.3.1.4

The expression needs to be subtracted from the dividend, so change all the signs in $x + 1$

				$1$
$x$	+	$1$		$x$	+	$0$
			-	$x$	-	$1$

Step 4.3.1.5

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

				$1$
$x$	+	$1$		$x$	+	$0$
			-	$x$	-	$1$
					-	$1$

Step 4.3.1.6

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

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

Step 4.3.2

Split the single integral into multiple integrals.

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

Step 4.3.3

Apply the constant rule.

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

Step 4.3.4

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

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

Step 4.3.5

Let $u = x + 1$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.3.5.1

Let $u = x + 1$ . Find $\frac{d u}{d x}$ .

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Step 4.3.5.1.1

Differentiate $x + 1$ .

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

Step 4.3.5.1.2

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

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

Step 4.3.5.1.3

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

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

Step 4.3.5.1.4

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

$1 + 0$

Step 4.3.5.1.5

Add $1$ and $0$ .

$1$

Step 4.3.5.2

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

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

Step 4.3.6

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

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

Step 4.3.7

Simplify.

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

Step 4.3.8

Replace all occurrences of $u$ with $x + 1$ .

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

Step 4.4

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

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