Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

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

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Step 1.1.1

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

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

Step 1.1.2

Simplify the left side.

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Step 1.1.2.1

Cancel the common factor of $y$ .

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Step 1.1.2.1.1

Cancel the common factor.

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

Step 1.1.2.1.2

Rewrite the expression.

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

Step 1.1.2.2

Cancel the common factor of $x + 1$ .

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Step 1.1.2.2.1

Cancel the common factor.

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

Step 1.1.2.2.2

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

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

Step 1.2

Regroup factors.

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

Step 1.3

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

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

Step 1.4

Simplify.

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Step 1.4.1

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

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

Step 1.4.2

Cancel the common factor of $y$ .

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Step 1.4.2.1

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

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

Step 1.4.2.2

Cancel the common factor.

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

Step 1.4.2.3

Rewrite the expression.

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

Step 1.4.3

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

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Step 1.4.3.1

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

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

Step 1.4.3.2

Cancel the common factor.

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

Step 1.4.3.3

Rewrite the expression.

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Differentiate $y^{2} + 1$ .

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

Step 2.2.1.1.2

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

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

Step 2.2.1.1.3

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

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

Step 2.2.1.1.4

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

$2 y + 0$

Step 2.2.1.1.5

Add $2 y$ and $0$ .

$2 y$

Step 2.2.1.2

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

$\int \frac{1}{u_{1}} \cdot \frac{1}{2} d u_{1} = \int \frac{x}{x + 1} d x$

Step 2.2.2

Simplify.

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Step 2.2.2.1

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

$\int \frac{1}{u_{1} \cdot 2} d u_{1} = \int \frac{x}{x + 1} d x$

Step 2.2.2.2

Move $2$ to the left of $u_{1}$ .

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Simplify.

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

Step 2.2.6

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Divide $x$ by $x + 1$ .

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Step 2.3.1.1

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 2.3.1.2

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

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

Step 2.3.1.3

Multiply the new quotient term by the divisor.

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

Step 2.3.1.4

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 2.3.1.5

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 2.3.1.6

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

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

Step 2.3.2

Split the single integral into multiple integrals.

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

Step 2.3.3

Apply the constant rule.

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

Step 2.3.4

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

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

Step 2.3.5

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

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Step 2.3.5.1

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

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Step 2.3.5.1.1

Differentiate $x + 1$ .

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

Step 2.3.5.1.2

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 2.3.5.1.3

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 2.3.5.1.4

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

$1 + 0$

Step 2.3.5.1.5

Add $1$ and $0$ .

$1$

Step 2.3.5.2

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

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

Step 2.3.6

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

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

Step 2.3.7

Simplify.

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

Step 2.3.8

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Multiply both sides of the equation by $2$ .

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

Step 3.2

Simplify both sides of the equation.

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

Simplify $2 (\frac{1}{2} \ln (| y^{2} + 1 |))$ .

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Step 3.2.1.1.1

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

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

Step 3.2.1.1.2

Cancel the common factor of $2$ .

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Step 3.2.1.1.2.1

Cancel the common factor.

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

Step 3.2.1.1.2.2

Rewrite the expression.

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

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $2 (x - \ln (| x + 1 |) + C)$ .

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Step 3.2.2.1.1

Apply the distributive property.

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

Step 3.2.2.1.2

Multiply $- 1$ by $2$ .

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

Step 3.3

Move all the terms containing a logarithm to the left side of the equation.

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

Step 3.4

Simplify the left side.

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Step 3.4.1

Simplify $\ln (| y^{2} + 1 |) + 2 \ln (| x + 1 |)$ .

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Step 3.4.1.1

Simplify each term.

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Step 3.4.1.1.1

Simplify $2 \ln (| x + 1 |)$ by moving $2$ inside the logarithm.

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

Step 3.4.1.1.2

Remove the absolute value in ${| x + 1 |}^{2}$ because exponentiations with even powers are always positive.

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

Step 3.4.1.2

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

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

Step 3.5

To solve for $y$ , rewrite the equation using properties of logarithms.

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

Step 3.6

Rewrite $\ln (| y^{2} + 1 | {(x + 1)}^{2}) = 2 x + 2 C$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{2 x + 2 C} = | y^{2} + 1 | {(x + 1)}^{2}$

Step 3.7

Solve for $y$ .

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Step 3.7.1

Rewrite the equation as $| y^{2} + 1 | {(x + 1)}^{2} = e^{2 x + 2 C}$ .

$| y^{2} + 1 | {(x + 1)}^{2} = e^{2 x + 2 C}$

Step 3.7.2

Divide each term in $| y^{2} + 1 | {(x + 1)}^{2} = e^{2 x + 2 C}$ by ${(x + 1)}^{2}$ and simplify.

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Step 3.7.2.1

Divide each term in $| y^{2} + 1 | {(x + 1)}^{2} = e^{2 x + 2 C}$ by ${(x + 1)}^{2}$ .

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

Step 3.7.2.2

Simplify the left side.

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Step 3.7.2.2.1

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

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Step 3.7.2.2.1.1

Cancel the common factor.

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

Step 3.7.2.2.1.2

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

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

Step 3.7.3

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$y^{2} + 1 = \pm \frac{e^{2 x + 2 C}}{{(x + 1)}^{2}}$

Step 3.7.4

Subtract $1$ from both sides of the equation.

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

Step 3.7.5

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{\pm \frac{e^{2 x + 2 C}}{{(x + 1)}^{2}} - 1}$

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

$y = \pm \sqrt{\pm \frac{e^{2 x + C}}{{(x + 1)}^{2}} - 1}$

Step 4.2

Rewrite $e^{2 x + C}$ as $e^{2 x} e^{C}$ .

$y = \pm \sqrt{\pm \frac{e^{2 x} e^{C}}{{(x + 1)}^{2}} - 1}$

Step 4.3

Reorder $e^{2 x}$ and $e^{C}$ .

$y = \pm \sqrt{\pm \frac{e^{C} e^{2 x}}{{(x + 1)}^{2}} - 1}$

Step 4.4

Combine constants with the plus or minus.

$y = \pm \sqrt{C \frac{C e^{2 x}}{{(x + 1)}^{2}} - 1}$