Calculus Examples

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

Step 1

Subtract $(1 + y^{2}) d x$ from both sides of the equation.

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

Rewrite using the commutative property of multiplication.

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

Cancel the common factor.

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

Step 3.2.3

Rewrite the expression.

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

Step 3.3

Move the negative in front of the fraction.

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

Step 3.4

Rewrite using the commutative property of multiplication.

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

Step 3.5

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

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Step 3.5.1

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

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

Step 3.5.2

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

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

Step 3.5.3

Cancel the common factor.

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

Step 3.5.4

Rewrite the expression.

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

Step 3.6

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

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

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

Step 4.2.2

Rewrite $1$ as $1^{2}$ .

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

Step 4.2.3

The integral of $\frac{1}{1^{2} + y^{2}}$ with respect to $y$ is $\arctan (y) + C_{1}$ .

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

Step 4.2.4

Simplify.

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

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.

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

Step 4.3.2

Rewrite $1$ as $1^{2}$ .

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

Step 4.3.3

The integral of $\frac{1}{1^{2} + x^{2}}$ with respect to $x$ is $\arctan (x) + C_{2}$ .

$- \arctan (y) + C_{1} = - (\arctan (x) + C_{2})$

Step 4.3.4

Simplify.

$- \arctan (y) + C_{1} = - \arctan (x) + C_{2}$

Step 4.4

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

$- \arctan (y) = - \arctan (x) + K$

Step 5

Solve for $y$ .

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Step 5.1

Divide each term in $- \arctan (y) = - \arctan (x) + K$ by $- 1$ and simplify.

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Step 5.1.1

Divide each term in $- \arctan (y) = - \arctan (x) + K$ by $- 1$ .

$\frac{- \arctan (y)}{- 1} = \frac{- \arctan (x)}{- 1} + \frac{K}{- 1}$

Step 5.1.2

Simplify the left side.

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Step 5.1.2.1

Dividing two negative values results in a positive value.

$\frac{\arctan (y)}{1} = \frac{- \arctan (x)}{- 1} + \frac{K}{- 1}$

Step 5.1.2.2

Divide $\arctan (y)$ by $1$ .

$\arctan (y) = \frac{- \arctan (x)}{- 1} + \frac{K}{- 1}$

Step 5.1.3

Simplify the right side.

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Step 5.1.3.1

Simplify each term.

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Step 5.1.3.1.1

Dividing two negative values results in a positive value.

$\arctan (y) = \frac{\arctan (x)}{1} + \frac{K}{- 1}$

Step 5.1.3.1.2

Divide $\arctan (x)$ by $1$ .

$\arctan (y) = \arctan (x) + \frac{K}{- 1}$

Step 5.1.3.1.3

Move the negative one from the denominator of $\frac{K}{- 1}$ .

$\arctan (y) = \arctan (x) - 1 \cdot K$

Step 5.1.3.1.4

Rewrite $- 1 \cdot K$ as $- K$ .

$\arctan (y) = \arctan (x) - K$

Step 5.2

Rewrite the equation as $\arctan (x) - K = \arctan (y)$ .

$\arctan (x) - K = \arctan (y)$

Step 5.3

Add $K$ to both sides of the equation.

$\arctan (x) = \arctan (y) + K$

Step 5.4

Take the inverse arctangent of both sides of the equation to extract $y$ from inside the arctangent.

$x = \tan (\arctan (y) + K)$

Step 5.5

Rewrite the equation as $\tan (\arctan (y) + K) = x$ .

$\tan (\arctan (y) + K) = x$

Step 5.6

Take the inverse tangent of both sides of the equation to extract $\arctan (y)$ from inside the tangent.

$\arctan (y) + K = \arctan (x)$

Step 5.7

Subtract $K$ from both sides of the equation.

$\arctan (y) = \arctan (x) - K$

Step 5.8

Take the inverse arctangent of both sides of the equation to extract $y$ from inside the arctangent.

$y = \tan (\arctan (x) - K)$

Step 5.9

Since $y$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$\tan (\arctan (y) + K) = x$

Step 5.10

Take the inverse tangent of both sides of the equation to extract $\arctan (y)$ from inside the tangent.

$\arctan (y) + K = \arctan (x)$

Step 5.11

Subtract $K$ from both sides of the equation.

$\arctan (y) = \arctan (x) - K$

Step 5.12

Take the inverse arctangent of both sides of the equation to extract $y$ from inside the arctangent.

$y = \tan (\arctan (x) - K)$

Step 6

Simplify the constant of integration.

$y = \tan (\arctan (x) + K)$