Calculus Examples

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

Step 1

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

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

Step 2

Simplify.

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

Cancel the common factor.

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

Step 2.1.3

Rewrite the expression.

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

Step 2.2

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

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

Step 2.3

Cancel the common factor of $y + 1$ .

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Step 2.3.1

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

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

Step 2.3.2

Cancel the common factor.

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

Step 2.3.3

Rewrite the expression.

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

Step 3

Integrate both sides.

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Step 3.1

Set up an integral on each side.

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

Step 3.2

Integrate the left side.

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Step 3.2.1

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

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Step 3.2.1.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$y$

$1$

$y^{2}$

$0 y$

$0$

Step 3.2.1.2

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

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

Step 3.2.1.3

Multiply the new quotient term by the divisor.

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

Step 3.2.1.4

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

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

Step 3.2.1.5

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

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

Step 3.2.1.6

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

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

Step 3.2.1.7

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

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

Step 3.2.1.8

Multiply the new quotient term by the divisor.

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

Step 3.2.1.9

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

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

Step 3.2.1.10

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

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

Step 3.2.1.11

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

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

Step 3.2.2

Split the single integral into multiple integrals.

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

Step 3.2.3

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

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

Step 3.2.4

Apply the constant rule.

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

Step 3.2.5

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

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Step 3.2.5.1

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

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Step 3.2.5.1.1

Differentiate $y + 1$ .

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

Step 3.2.5.1.2

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 3.2.5.1.3

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 3.2.5.1.4

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

$1 + 0$

Step 3.2.5.1.5

Add $1$ and $0$ .

$1$

Step 3.2.5.2

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

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

Step 3.2.6

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

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

Step 3.2.7

Simplify.

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

Step 3.2.8

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

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

Step 3.3

Integrate the right side.

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Step 3.3.1

Apply basic rules of exponents.

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Step 3.3.1.1

Move $x^{2}$ out of the denominator by raising it to the $- 1$ power.

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

Step 3.3.1.2

Multiply the exponents in ${(x^{2})}^{- 1}$ .

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Step 3.3.1.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 3.3.1.2.2

Multiply $2$ by $- 1$ .

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

Step 3.3.2

By the Power Rule, the integral of $x^{- 2}$ with respect to $x$ is $- x^{- 1}$ .

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

Step 3.3.3

Rewrite $- x^{- 1} + C_{5}$ as $- \frac{1}{x} + C_{5}$ .

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

Step 3.4

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

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