Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Regroup factors.

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

Step 1.2

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

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

Step 1.3

Cancel the common factor of $y$ .

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Step 1.3.1

Factor $y$ out of $\frac{x}{x + 1} y$ .

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

Step 1.3.2

Cancel the common factor.

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

Step 1.3.3

Rewrite the expression.

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

Step 1.4

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

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

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

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.

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

Step 2.3.2

Split the single integral into multiple integrals.

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

Step 2.3.3

Apply the constant rule.

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

Step 2.3.4

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

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

Step 2.3.5

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

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Step 2.3.5.1

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

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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 $u$ and $d u$ .

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

Step 2.3.6

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

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

Step 2.3.7

Simplify.

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

Step 2.3.8

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

To multiply absolute values, multiply the terms inside each absolute value.

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

Step 3.4

Apply the distributive property.

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

Step 3.5

Multiply $y$ by $1$ .

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

Step 3.6

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

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

Step 3.7

Rewrite $\ln (| y x + y |) = x + C$ 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 $b^{y} = x$ .

$e^{x + C} = | y x + y |$

Step 3.8

Solve for $y$ .

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Step 3.8.1

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

$| y x + y | = e^{x + C}$

Step 3.8.2

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

$y x + y = \pm e^{x + C}$

Step 3.8.3

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

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Step 3.8.3.1

Factor $y$ out of $y x$ .

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

Step 3.8.3.2

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

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

Step 3.8.3.3

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

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

Step 3.8.3.4

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

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

Step 3.8.4

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

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Step 3.8.4.1

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

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

Step 3.8.4.2

Simplify the left side.

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Step 3.8.4.2.1

Cancel the common factor of $x + 1$ .

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Step 3.8.4.2.1.1

Cancel the common factor.

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

Step 3.8.4.2.1.2

Divide $y$ by $1$ .

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

Step 4

Group the constant terms together.

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

Combine constants with the plus or minus.

$y = \frac{C e^{x}}{x + 1}$