Calculus Examples

$x (2 y - 3) d x + (x^{2} + 1) d y = 0$

Step 1

Subtract $x (2 y - 3) d x$ from both sides of the equation.

$(x^{2} + 1) d y = - x (2 y - 3) d x$

Step 2

Multiply both sides by $\frac{1}{(x^{2} + 1) (2 y - 3)}$ .

$\frac{1}{(x^{2} + 1) (2 y - 3)} (x^{2} + 1) d y = \frac{1}{(x^{2} + 1) (2 y - 3)} (- x (2 y - 3)) d x$

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

Cancel the common factor.

$\frac{1}{(x^{2} + 1) (2 y - 3)} (x^{2} + 1) d y = \frac{1}{(x^{2} + 1) (2 y - 3)} (- x (2 y - 3)) d x$

Step 3.1.2

Rewrite the expression.

$\frac{1}{2 y - 3} d y = \frac{1}{(x^{2} + 1) (2 y - 3)} (- x (2 y - 3)) d x$

Step 3.2

Rewrite using the commutative property of multiplication.

$\frac{1}{2 y - 3} d y = - \frac{1}{(x^{2} + 1) (2 y - 3)} (x (2 y - 3)) d x$

Step 3.3

Cancel the common factor of $2 y - 3$ .

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Step 3.3.1

Move the leading negative in $- \frac{1}{(x^{2} + 1) (2 y - 3)}$ into the numerator.

$\frac{1}{2 y - 3} d y = \frac{- 1}{(x^{2} + 1) (2 y - 3)} (x (2 y - 3)) d x$

Step 3.3.2

Factor $2 y - 3$ out of $(x^{2} + 1) (2 y - 3)$ .

$\frac{1}{2 y - 3} d y = \frac{- 1}{(2 y - 3) (x^{2} + 1)} (x (2 y - 3)) d x$

Step 3.3.3

Factor $2 y - 3$ out of $x (2 y - 3)$ .

$\frac{1}{2 y - 3} d y = \frac{- 1}{(2 y - 3) (x^{2} + 1)} ((2 y - 3) x) d x$

Step 3.3.4

Cancel the common factor.

$\frac{1}{2 y - 3} d y = \frac{- 1}{(2 y - 3) (x^{2} + 1)} ((2 y - 3) x) d x$

Step 3.3.5

Rewrite the expression.

$\frac{1}{2 y - 3} d y = \frac{- 1}{x^{2} + 1} x d x$

Step 3.4

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

$\frac{1}{2 y - 3} d y = \frac{- x}{x^{2} + 1} d x$

Step 3.5

Move the negative in front of the fraction.

$\frac{1}{2 y - 3} d y = - \frac{x}{x^{2} + 1} d x$

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

$\int \frac{1}{2 y - 3} d y = \int - \frac{x}{x^{2} + 1} d x$

Step 4.2

Integrate the left side.

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Step 4.2.1

Let $u_{1} = 2 y - 3$ . Then $d u_{1} = 2 d y$ , so $\frac{1}{2} d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 4.2.1.1

Let $u_{1} = 2 y - 3$ . Find $\frac{d u_{1}}{d y}$ .

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Step 4.2.1.1.1

Differentiate $2 y - 3$ .

$\frac{d}{d y} [2 y - 3]$

Step 4.2.1.1.2

By the Sum Rule, the derivative of $2 y - 3$ with respect to $y$ is $\frac{d}{d y} [2 y] + \frac{d}{d y} [- 3]$ .

$\frac{d}{d y} [2 y] + \frac{d}{d y} [- 3]$

Step 4.2.1.1.3

Evaluate $\frac{d}{d y} [2 y]$ .

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Step 4.2.1.1.3.1

Since $2$ is constant with respect to $y$ , the derivative of $2 y$ with respect to $y$ is $2 \frac{d}{d y} [y]$ .

$2 \frac{d}{d y} [y] + \frac{d}{d y} [- 3]$

Step 4.2.1.1.3.2

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

$2 \cdot 1 + \frac{d}{d y} [- 3]$

Step 4.2.1.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d y} [- 3]$

Step 4.2.1.1.4

Differentiate using the Constant Rule.

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Step 4.2.1.1.4.1

Since $- 3$ is constant with respect to $y$ , the derivative of $- 3$ with respect to $y$ is $0$ .

$2 + 0$

Step 4.2.1.1.4.2

Add $2$ and $0$ .

$2$

Step 4.2.1.2

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

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

Step 4.2.2

Simplify.

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Step 4.2.2.1

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

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

Step 4.2.2.2

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

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

Step 4.2.3

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

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

Step 4.2.4

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

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

Step 4.2.5

Simplify.

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

Step 4.2.6

Replace all occurrences of $u_{1}$ with $2 y - 3$ .

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

Step 4.3

Integrate the right side.

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Step 4.3.1

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

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

Step 4.3.2

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 4.3.2.1

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

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Step 4.3.2.1.1

Differentiate $x^{2} + 1$ .

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

Step 4.3.2.1.2

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 4.3.2.1.3

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 4.3.2.1.4

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

$2 x + 0$

Step 4.3.2.1.5

Add $2 x$ and $0$ .

$2 x$

Step 4.3.2.2

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

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

Step 4.3.3

Simplify.

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Step 4.3.3.1

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

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

Step 4.3.3.2

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

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

Step 4.3.4

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

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

Step 4.3.5

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

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

Step 4.3.6

Simplify.

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

Step 4.3.7

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

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

Step 4.4

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

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

Step 5

Solve for $y$ .

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Step 5.1

Multiply both sides of the equation by $2$ .

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

Step 5.2

Simplify both sides of the equation.

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Step 5.2.1

Simplify the left side.

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Step 5.2.1.1

Simplify $2 (\frac{1}{2} \ln (| 2 y - 3 |))$ .

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Step 5.2.1.1.1

Combine $\frac{1}{2}$ and $\ln (| 2 y - 3 |)$ .

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

Step 5.2.1.1.2

Cancel the common factor of $2$ .

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Step 5.2.1.1.2.1

Cancel the common factor.

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

Step 5.2.1.1.2.2

Rewrite the expression.

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

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

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

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Step 5.2.2.1.1

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

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

Step 5.2.2.1.2

To write $C$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 5.2.2.1.3

Simplify terms.

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Step 5.2.2.1.3.1

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

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

Step 5.2.2.1.3.2

Combine the numerators over the common denominator.

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

Step 5.2.2.1.3.3

Cancel the common factor of $2$ .

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Step 5.2.2.1.3.3.1

Cancel the common factor.

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

Step 5.2.2.1.3.3.2

Rewrite the expression.

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

Step 5.2.2.1.4

Move $2$ to the left of $C$ .

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

Step 5.3

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

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

Step 5.4

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

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

Step 5.5

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

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

Step 5.6

Expand $(2 y - 3) (x^{2} + 1)$ using the FOIL Method.

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Step 5.6.1

Apply the distributive property.

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

Step 5.6.2

Apply the distributive property.

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

Step 5.6.3

Apply the distributive property.

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

Step 5.7

Simplify each term.

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Step 5.7.1

Multiply $2$ by $1$ .

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

Step 5.7.2

Multiply $- 3$ by $1$ .

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

Step 5.8

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

$e^{\ln (| 2 y x^{2} + 2 y - 3 x^{2} - 3 |)} = e^{2 C}$

Step 5.9

Rewrite $\ln (| 2 y x^{2} + 2 y - 3 x^{2} - 3 |) = 2 C$ 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 $b^{y} = x$ .

$e^{2 C} = | 2 y x^{2} + 2 y - 3 x^{2} - 3 |$

Step 5.10

Solve for $y$ .

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Step 5.10.1

Rewrite the equation as $| 2 y x^{2} + 2 y - 3 x^{2} - 3 | = e^{2 C}$ .

$| 2 y x^{2} + 2 y - 3 x^{2} - 3 | = e^{2 C}$

Step 5.10.2

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

$2 y x^{2} + 2 y - 3 x^{2} - 3 = \pm e^{2 C}$

Step 5.10.3

Move all terms not containing $y$ to the right side of the equation.

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Step 5.10.3.1

Add $3 x^{2}$ to both sides of the equation.

$2 y x^{2} + 2 y - 3 = \pm e^{2 C} + 3 x^{2}$

Step 5.10.3.2

Add $3$ to both sides of the equation.

$2 y x^{2} + 2 y = \pm e^{2 C} + 3 x^{2} + 3$

Step 5.10.4

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

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Step 5.10.4.1

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

$2 y (x^{2}) + 2 y = \pm e^{2 C} + 3 x^{2} + 3$

Step 5.10.4.2

Factor $2 y$ out of $2 y$ .

$2 y (x^{2}) + 2 y (1) = \pm e^{2 C} + 3 x^{2} + 3$

Step 5.10.4.3

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

$2 y (x^{2} + 1) = \pm e^{2 C} + 3 x^{2} + 3$

Step 5.10.5

Divide each term in $2 y (x^{2} + 1) = \pm e^{2 C} + 3 x^{2} + 3$ by $2 (x^{2} + 1)$ and simplify.

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Step 5.10.5.1

Divide each term in $2 y (x^{2} + 1) = \pm e^{2 C} + 3 x^{2} + 3$ by $2 (x^{2} + 1)$ .

$\frac{2 y (x^{2} + 1)}{2 (x^{2} + 1)} = \frac{\pm e^{2 C}}{2 (x^{2} + 1)} + \frac{3 x^{2}}{2 (x^{2} + 1)} + \frac{3}{2 (x^{2} + 1)}$

Step 5.10.5.2

Simplify the left side.

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Step 5.10.5.2.1

Cancel the common factor of $2$ .

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Step 5.10.5.2.1.1

Cancel the common factor.

$\frac{2 y (x^{2} + 1)}{2 (x^{2} + 1)} = \frac{\pm e^{2 C}}{2 (x^{2} + 1)} + \frac{3 x^{2}}{2 (x^{2} + 1)} + \frac{3}{2 (x^{2} + 1)}$

Step 5.10.5.2.1.2

Rewrite the expression.

$\frac{y (x^{2} + 1)}{x^{2} + 1} = \frac{\pm e^{2 C}}{2 (x^{2} + 1)} + \frac{3 x^{2}}{2 (x^{2} + 1)} + \frac{3}{2 (x^{2} + 1)}$

Step 5.10.5.2.2

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

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Step 5.10.5.2.2.1

Cancel the common factor.

$\frac{y (x^{2} + 1)}{x^{2} + 1} = \frac{\pm e^{2 C}}{2 (x^{2} + 1)} + \frac{3 x^{2}}{2 (x^{2} + 1)} + \frac{3}{2 (x^{2} + 1)}$

Step 5.10.5.2.2.2

Divide $y$ by $1$ .

$y = \frac{\pm e^{2 C}}{2 (x^{2} + 1)} + \frac{3 x^{2}}{2 (x^{2} + 1)} + \frac{3}{2 (x^{2} + 1)}$

Step 6

Simplify the constant of integration.

$y = \frac{C}{2 (x^{2} + 1)} + \frac{3 x^{2}}{2 (x^{2} + 1)} + \frac{3}{2 (x^{2} + 1)}$