Calculus Examples

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

Step 1

Rewrite the equation.

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

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

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Step 3.1.1

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

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

Step 3.1.2

Cancel the common factor.

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

Step 3.1.3

Rewrite the expression.

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

Step 3.2

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

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

Step 3.3

Cancel the common factor of $y + 1$ .

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

Cancel the common factor.

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

Step 3.3.4

Rewrite the expression.

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

Step 3.4

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

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

Divide $y^{2} + 1$ 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 $0$ .

$y$

$1$

$y^{2}$

$0 y$

$1$

Step 4.2.1.2

Divide the highest order term in the dividend $y^{2}$ by the highest order term in divisor $y$ .

					$y$
	$y$	+	$1$		$y^{2}$	+	$0 y$	+	$1$

Step 4.2.1.3

Multiply the new quotient term by the divisor.

				$y$
$y$	+	$1$		$y^{2}$	+	$0 y$	+	$1$
			+	$y^{2}$	+	$y$

Step 4.2.1.4

The expression needs to be subtracted from the dividend, so change all the signs in $y^{2} + y$

				$y$
$y$	+	$1$		$y^{2}$	+	$0 y$	+	$1$
			-	$y^{2}$	-	$y$

Step 4.2.1.5

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

				$y$
$y$	+	$1$		$y^{2}$	+	$0 y$	+	$1$
			-	$y^{2}$	-	$y$
					-	$y$

Step 4.2.1.6

Pull the next terms from the original dividend down into the current dividend.

				$y$
$y$	+	$1$		$y^{2}$	+	$0 y$	+	$1$
			-	$y^{2}$	-	$y$
					-	$y$	+	$1$

Step 4.2.1.7

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

				$y$	-	$1$
$y$	+	$1$		$y^{2}$	+	$0 y$	+	$1$
			-	$y^{2}$	-	$y$
					-	$y$	+	$1$

Step 4.2.1.8

Multiply the new quotient term by the divisor.

				$y$	-	$1$
$y$	+	$1$		$y^{2}$	+	$0 y$	+	$1$
			-	$y^{2}$	-	$y$
					-	$y$	+	$1$
					-	$y$	-	$1$

Step 4.2.1.9

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

				$y$	-	$1$
$y$	+	$1$		$y^{2}$	+	$0 y$	+	$1$
			-	$y^{2}$	-	$y$
					-	$y$	+	$1$
					+	$y$	+	$1$

Step 4.2.1.10

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

				$y$	-	$1$
$y$	+	$1$		$y^{2}$	+	$0 y$	+	$1$
			-	$y^{2}$	-	$y$
					-	$y$	+	$1$
					+	$y$	+	$1$
							+	$2$

Step 4.2.1.11

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

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

Step 4.2.2

Split the single integral into multiple integrals.

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

Step 4.2.3

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

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

Step 4.2.4

Apply the constant rule.

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

Step 4.2.5

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

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

Step 4.2.6

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

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Step 4.2.6.1

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

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Step 4.2.6.1.1

Differentiate $y + 1$ .

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

Step 4.2.6.1.2

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 4.2.6.1.3

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 4.2.6.1.4

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

$1 + 0$

Step 4.2.6.1.5

Add $1$ and $0$ .

$1$

Step 4.2.6.2

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

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

Step 4.2.7

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

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

Step 4.2.8

Simplify.

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

Step 4.2.9

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

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

Step 4.3

Integrate the right side.

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Step 4.3.1

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 4.3.1.1

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

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Step 4.3.1.1.1

Differentiate $x^{2} + 1$ .

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

Step 4.3.1.1.2

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 4.3.1.1.3

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 4.3.1.1.4

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

$2 x + 0$

Step 4.3.1.1.5

Add $2 x$ and $0$ .

$2 x$

Step 4.3.1.2

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

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

Step 4.3.2

Simplify.

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Step 4.3.2.1

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

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

Step 4.3.2.2

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

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

Step 4.3.3

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

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

Step 4.3.4

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

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

Step 4.3.5

Simplify.

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

Step 4.3.6

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

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

Step 4.4

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

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