Calculus Examples

$4 x y^{2} d x + (x^{2} + 1) d y = 0$

Step 1

Subtract $4 x y^{2} d x$ from both sides of the equation.

$(x^{2} + 1) d y = - 4 x y^{2} d x$

Step 2

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

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

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

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

Step 3.1.2

Cancel the common factor.

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

Step 3.1.3

Rewrite the expression.

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

Step 3.2

Rewrite using the commutative property of multiplication.

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

Step 3.3

Combine $- 4$ and $\frac{1}{(x^{2} + 1) y^{2}}$ .

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

Step 3.4

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

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Step 3.4.1

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

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

Step 3.4.2

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

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

Step 3.4.3

Cancel the common factor.

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

Step 3.4.4

Rewrite the expression.

$\frac{1}{y^{2}} d y = \frac{- 4}{x^{2} + 1} x d x$

Step 3.5

Combine $\frac{- 4}{x^{2} + 1}$ and $x$ .

$\frac{1}{y^{2}} d y = \frac{- 4 x}{x^{2} + 1} d x$

Step 3.6

Move the negative in front of the fraction.

$\frac{1}{y^{2}} d y = - \frac{4 x}{x^{2} + 1} d x$

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

$\int \frac{1}{y^{2}} d y = \int - \frac{4 x}{x^{2} + 1} d x$

Step 4.2

Integrate the left side.

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Step 4.2.1

Apply basic rules of exponents.

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Step 4.2.1.1

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

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

Step 4.2.1.2

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

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Step 4.2.1.2.1

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

$\int y^{2 \cdot - 1} d y = \int - \frac{4 x}{x^{2} + 1} d x$

Step 4.2.1.2.2

Multiply $2$ by $- 1$ .

$\int y^{- 2} d y = \int - \frac{4 x}{x^{2} + 1} d x$

Step 4.2.2

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

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

Step 4.2.3

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

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

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.

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

Step 4.3.2

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

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

Step 4.3.3

Multiply $4$ by $- 1$ .

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

Step 4.3.4

Let $u = x^{2} + 1$ . Then $d u = 2 x d x$ , so $\frac{1}{2} d u = x d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.3.4.1

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

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Step 4.3.4.1.1

Differentiate $x^{2} + 1$ .

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

Step 4.3.4.1.2

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 4.3.4.1.3

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 4.3.4.1.4

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

$2 x + 0$

Step 4.3.4.1.5

Add $2 x$ and $0$ .

$2 x$

Step 4.3.4.2

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

$- \frac{1}{y} + C_{1} = - 4 \int \frac{1}{u} \cdot \frac{1}{2} d u$

Step 4.3.5

Simplify.

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Step 4.3.5.1

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

$- \frac{1}{y} + C_{1} = - 4 \int \frac{1}{u \cdot 2} d u$

Step 4.3.5.2

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

$- \frac{1}{y} + C_{1} = - 4 \int \frac{1}{2 u} d u$

Step 4.3.6

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

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

Step 4.3.7

Simplify.

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Step 4.3.7.1

Combine $\frac{1}{2}$ and $- 4$ .

$- \frac{1}{y} + C_{1} = \frac{- 4}{2} \int \frac{1}{u} d u$

Step 4.3.7.2

Cancel the common factor of $- 4$ and $2$ .

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Step 4.3.7.2.1

Factor $2$ out of $- 4$ .

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

Step 4.3.7.2.2

Cancel the common factors.

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Step 4.3.7.2.2.1

Factor $2$ out of $2$ .

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

Step 4.3.7.2.2.2

Cancel the common factor.

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

Step 4.3.7.2.2.3

Rewrite the expression.

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

Step 4.3.7.2.2.4

Divide $- 2$ by $1$ .

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

Step 4.3.8

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

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

Step 4.3.9

Simplify.

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

Step 4.3.10

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

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

Step 4.4

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

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

Step 5

Solve for $y$ .

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Step 5.1

Simplify $- 2 \ln (| x^{2} + 1 |)$ by moving $2$ inside the logarithm.

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

Step 5.2

Find the LCD of the terms in the equation.

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Step 5.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$y, 1, 1$

Step 5.2.2

The LCM of one and any expression is the expression.

$y$

Step 5.3

Multiply each term in $- \frac{1}{y} = - \ln ({| x^{2} + 1 |}^{2}) + C$ by $y$ to eliminate the fractions.

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Step 5.3.1

Multiply each term in $- \frac{1}{y} = - \ln ({| x^{2} + 1 |}^{2}) + C$ by $y$ .

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $y$ .

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Step 5.3.2.1.1

Move the leading negative in $- \frac{1}{y}$ into the numerator.

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

Step 5.3.2.1.2

Cancel the common factor.

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

Step 5.3.2.1.3

Rewrite the expression.

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

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Remove the absolute value in ${| x^{2} + 1 |}^{2}$ because exponentiations with even powers are always positive.

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

Step 5.3.3.2

Reorder factors in $- \ln ({(x^{2} + 1)}^{2}) y + C y$ .

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

Step 5.4

Solve the equation.

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Step 5.4.1

Rewrite the equation as $- y \ln ({(x^{2} + 1)}^{2}) + C y = - 1$ .

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

Step 5.4.2

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

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Step 5.4.2.1

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

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

Step 5.4.2.2

Factor $y$ out of $C y$ .

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

Step 5.4.2.3

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

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

Step 5.4.3

Rewrite $- 1 \ln ({(x^{2} + 1)}^{2})$ as $- \ln ({(x^{2} + 1)}^{2})$ .

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

Step 5.4.4

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

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Step 5.4.4.1

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

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

Step 5.4.4.2

Simplify the left side.

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Step 5.4.4.2.1

Cancel the common factor of $- \ln ({(x^{2} + 1)}^{2}) + C$ .

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Step 5.4.4.2.1.1

Cancel the common factor.

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

Step 5.4.4.2.1.2

Divide $y$ by $1$ .

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

Step 5.4.4.3

Simplify the right side.

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Step 5.4.4.3.1

Move the negative in front of the fraction.

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

Step 5.4.4.3.2

Factor $- 1$ out of $- \ln ({(x^{2} + 1)}^{2})$ .

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

Step 5.4.4.3.3

Factor $- 1$ out of $C$ .

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

Step 5.4.4.3.4

Factor $- 1$ out of $- (\ln ({(x^{2} + 1)}^{2})) - 1 (- C)$ .

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

Step 5.4.4.3.5

Simplify the expression.

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Step 5.4.4.3.5.1

Rewrite $- (\ln ({(x^{2} + 1)}^{2}) - C)$ as $- 1 (\ln ({(x^{2} + 1)}^{2}) - C)$ .

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

Step 5.4.4.3.5.2

Move the negative in front of the fraction.

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

Step 5.4.4.3.5.3

Multiply $- 1$ by $- 1$ .

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

Step 5.4.4.3.5.4

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

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

Step 6

Simplify the constant of integration.

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