Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

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

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

Step 1.2

Simplify.

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Step 1.2.1

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

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Step 1.2.1.1

Cancel the common factor.

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

Step 1.2.1.2

Rewrite the expression.

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

Step 1.2.2

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

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Step 1.2.2.1

Factor $2$ out of $2$ .

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

Step 1.2.2.2

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

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

Step 1.2.2.3

Factor $2$ out of $2 (1) + 2 (- y)$ .

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Simplify the expression.

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Step 2.2.1.1

Reorder $x^{2}$ and $1$ .

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

Step 2.2.1.2

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

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

Step 2.2.2

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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

Step 2.3.2

Let $u = 1 - y$ . Then $d u = - d y$ , so $- d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.2.1

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

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Step 2.3.2.1.1

Rewrite.

$\frac{- 1}{1}$

Step 2.3.2.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 2.3.2.2

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

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

Step 2.3.3

Move the negative in front of the fraction.

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

Step 2.3.4

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

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

Step 2.3.5

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

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

Step 2.3.6

Simplify.

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Step 2.3.6.1

Simplify.

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

Step 2.3.6.2

Combine $\frac{1}{2}$ and $\ln (| u |)$ .

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

Step 2.3.7

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

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

Step 2.3.8

Reorder terms.

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

Step 2.4

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

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

Step 3

Solve for $x$ .

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Step 3.1

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

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

Step 3.2

Simplify the right side.

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Step 3.2.1

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

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Step 3.2.1.1

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

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

Step 3.2.1.2

Simplify $\frac{1}{2} \ln (| 1 - y |)$ by moving $\frac{1}{2}$ inside the logarithm.

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