Calculus Examples

$y (x - 1) d y - x (y - 1) d x = 0$

Step 1

Add $x (y - 1) d x$ to both sides of the equation.

$y (x - 1) d y = x (y - 1) d x$

Step 2

Multiply both sides by $\frac{1}{(x - 1) (y - 1)}$ .

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

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 $y (x - 1)$ .

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

Step 3.1.2

Cancel the common factor.

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

Step 3.1.3

Rewrite the expression.

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

Step 3.2

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

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

Step 3.3

Cancel the common factor of $y - 1$ .

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Step 3.3.1

Factor $y - 1$ out of $(x - 1) (y - 1)$ .

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

Step 3.3.2

Factor $y - 1$ out of $x (y - 1)$ .

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

Step 3.3.3

Cancel the common factor.

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

Step 3.3.4

Rewrite the expression.

$\frac{y}{y - 1} d y = \frac{1}{x - 1} x d x$

Step 3.4

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

$\frac{y}{y - 1} d y = \frac{x}{x - 1} d x$

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

Divide $y$ by $y - 1$ .

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Step 4.2.1.1

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

$y$

$1$

$y$

$0$

Step 4.2.1.2

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

					$1$
	$y$	-	$1$		$y$	+	$0$

Step 4.2.1.3

Multiply the new quotient term by the divisor.

				$1$
$y$	-	$1$		$y$	+	$0$
			+	$y$	-	$1$

Step 4.2.1.4

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

				$1$
$y$	-	$1$		$y$	+	$0$
			-	$y$	+	$1$

Step 4.2.1.5

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

				$1$
$y$	-	$1$		$y$	+	$0$
			-	$y$	+	$1$
					+	$1$

Step 4.2.1.6

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

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

Step 4.2.2

Split the single integral into multiple integrals.

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

Step 4.2.3

Apply the constant rule.

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

Step 4.2.4

Let $u_{1} = y - 1$ . Then $d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 4.2.4.1

Let $u_{1} = y - 1$ . Find $\frac{d u_{1}}{d y}$ .

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Step 4.2.4.1.1

Differentiate $y - 1$ .

$\frac{d}{d y} [y - 1]$

Step 4.2.4.1.2

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

$\frac{d}{d y} [y] + \frac{d}{d y} [- 1]$

Step 4.2.4.1.3

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

$1 + \frac{d}{d y} [- 1]$

Step 4.2.4.1.4

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

$1 + 0$

Step 4.2.4.1.5

Add $1$ and $0$ .

$1$

Step 4.2.4.2

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

$y + C_{1} + \int \frac{1}{u_{1}} d u_{1} = \int \frac{x}{x - 1} d x$

Step 4.2.5

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

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

Step 4.2.6

Simplify.

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

Step 4.2.7

Replace all occurrences of $u_{1}$ with $y - 1$ .

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

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.

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

Step 4.3.2

Split the single integral into multiple integrals.

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

Step 4.3.3

Apply the constant rule.

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

Step 4.3.4

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

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Step 4.3.4.1

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

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Step 4.3.4.1.1

Differentiate $x - 1$ .

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

Step 4.3.4.1.2

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 4.3.4.1.3

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 4.3.4.1.4

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

$1 + 0$

Step 4.3.4.1.5

Add $1$ and $0$ .

$1$

Step 4.3.4.2

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

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

Step 4.3.5

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

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

Step 4.3.6

Simplify.

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

Step 4.3.7

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

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

Step 4.4

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

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