Calculus Examples

$2 y (x + 1) d y = x d x$

Step 1

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

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

Step 2

Simplify.

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Step 2.1

Rewrite using the commutative property of multiplication.

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

Step 2.2

Combine $2$ and $\frac{1}{x + 1}$ .

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

Step 2.3

Cancel the common factor of $x + 1$ .

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Step 2.3.1

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

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

Step 2.3.2

Cancel the common factor.

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

Step 2.3.3

Rewrite the expression.

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

Step 2.4

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

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

Step 3

Integrate both sides.

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Step 3.1

Set up an integral on each side.

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

Step 3.2

Integrate the left side.

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Step 3.2.1

Since $2$ is constant with respect to $y$ , move $2$ out of the integral.

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

Step 3.2.2

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

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

Step 3.2.3

Simplify the answer.

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Step 3.2.3.1

Rewrite $2 (\frac{1}{2} y^{2} + C_{1})$ as $2 (\frac{1}{2}) y^{2} + C_{1}$ .

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

Step 3.2.3.2

Simplify.

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Step 3.2.3.2.1

Combine $2$ and $\frac{1}{2}$ .

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

Step 3.2.3.2.2

Cancel the common factor of $2$ .

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Step 3.2.3.2.2.1

Cancel the common factor.

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

Step 3.2.3.2.2.2

Rewrite the expression.

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

Step 3.2.3.2.3

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

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

Step 3.3

Integrate the right side.

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Step 3.3.1

Divide $x$ by $x + 1$ .

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Step 3.3.1.1

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 3.3.1.2

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

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

Step 3.3.1.3

Multiply the new quotient term by the divisor.

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

Step 3.3.1.4

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 3.3.1.5

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 3.3.1.6

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

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

Step 3.3.2

Split the single integral into multiple integrals.

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

Step 3.3.3

Apply the constant rule.

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

Step 3.3.4

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

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

Step 3.3.5

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

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Step 3.3.5.1

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

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Step 3.3.5.1.1

Differentiate $x + 1$ .

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

Step 3.3.5.1.2

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 3.3.5.1.3

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 3.3.5.1.4

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

$1 + 0$

Step 3.3.5.1.5

Add $1$ and $0$ .

$1$

Step 3.3.5.2

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

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

Step 3.3.6

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

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

Step 3.3.7

Simplify.

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

Step 3.3.8

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

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

Step 3.4

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

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

Step 4

Solve for $y$ .

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Step 4.1

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

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

Step 4.2

The complete solution is the result of both the positive and negative portions of the solution.

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Step 4.2.1

First, use the positive value of the $\pm$ to find the first solution.

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

Step 4.2.2

Next, use the negative value of the $\pm$ to find the second solution.

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

Step 4.2.3

The complete solution is the result of both the positive and negative portions of the solution.

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

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

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

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

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

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