Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Factor $x$ out of $x y + x$ .

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Step 1.1.1

Factor $x$ out of $x y$ .

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

Step 1.1.2

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

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

Step 1.1.3

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

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

Step 1.1.4

Factor $x$ out of $x (y) + x \cdot 1$ .

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

Step 1.2

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

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.2.4

Factor $y$ out of $y x^{2} + y \cdot 1$ .

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

Step 1.3

Regroup factors.

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

Step 1.4

Multiply both sides by $\frac{y}{y + 1}$ .

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

Step 1.5

Simplify.

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Step 1.5.1

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

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

Step 1.5.2

Cancel the common factor of $y$ .

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Step 1.5.2.1

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

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

Step 1.5.2.2

Cancel the common factor.

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

Step 1.5.2.3

Rewrite the expression.

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

Step 1.5.3

Cancel the common factor of $y + 1$ .

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Step 1.5.3.1

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

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

Step 1.5.3.2

Cancel the common factor.

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

Step 1.5.3.3

Rewrite the expression.

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

Step 1.6

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Divide $y$ by $y + 1$ .

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Step 2.2.1.1

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 2.2.1.2

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

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

Step 2.2.1.3

Multiply the new quotient term by the divisor.

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

Step 2.2.1.4

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 2.2.1.5

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 2.2.1.6

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

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

Step 2.2.2

Split the single integral into multiple integrals.

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

Step 2.2.3

Apply the constant rule.

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

Step 2.2.4

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

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

Step 2.2.5

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

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Step 2.2.5.1

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

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Step 2.2.5.1.1

Differentiate $y + 1$ .

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

Step 2.2.5.1.2

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 2.2.5.1.3

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 2.2.5.1.4

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

$1 + 0$

Step 2.2.5.1.5

Add $1$ and $0$ .

$1$

Step 2.2.5.2

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

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

Step 2.2.6

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

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

Step 2.2.7

Simplify.

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

Step 2.2.8

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Let $u_{2} = x^{2} + 1$ . Then $d u_{2} = 2 x d x$ , so $\frac{1}{2} d u_{2} = x d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 2.3.1.1

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

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Step 2.3.1.1.1

Differentiate $x^{2} + 1$ .

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

Step 2.3.1.1.2

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

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

Step 2.3.1.1.3

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

$2 x + \frac{d}{d x} [1]$

Step 2.3.1.1.4

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

$2 x + 0$

Step 2.3.1.1.5

Add $2 x$ and $0$ .

$2 x$

Step 2.3.1.2

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

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

Step 2.3.2

Simplify.

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Step 2.3.2.1

Multiply $\frac{1}{u_{2}}$ by $\frac{1}{2}$ .

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

Step 2.3.2.2

Move $2$ to the left of $u_{2}$ .

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

Step 2.3.3

Since $\frac{1}{2}$ is constant with respect to $u_{2}$ , move $\frac{1}{2}$ out of the integral.

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

Step 2.3.4

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

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

Step 2.3.5

Simplify.

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

Step 2.3.6

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

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

Step 2.4

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

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