Calculus Examples

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

Step 1

Add $2 x (1 + e^{y}) d x$ to both sides of the equation.

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

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

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Step 3.1.1

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

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

Step 3.1.2

Cancel the common factor.

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

Step 3.1.3

Rewrite the expression.

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

Step 3.2

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

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

Step 3.3

Rewrite using the commutative property of multiplication.

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

Step 3.4

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

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

Step 3.5

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

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Step 3.5.1

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

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

Step 3.5.2

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

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

Step 3.5.3

Cancel the common factor.

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

Step 3.5.4

Rewrite the expression.

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

Step 3.6

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

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

Let $u_{1} = 1 + e^{y}$ . Then $d u_{1} = e^{y} d y$ , so $\frac{1}{e^{y}} d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 4.2.1.1

Let $u_{1} = 1 + e^{y}$ . Find $\frac{d u_{1}}{d y}$ .

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Step 4.2.1.1.1

Differentiate $1 + e^{y}$ .

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

Step 4.2.1.1.2

Differentiate.

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Step 4.2.1.1.2.1

By the Sum Rule, the derivative of $1 + e^{y}$ with respect to $y$ is $\frac{d}{d y} [1] + \frac{d}{d y} [e^{y}]$ .

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

Step 4.2.1.1.2.2

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

$0 + \frac{d}{d y} [e^{y}]$

Step 4.2.1.1.3

Differentiate using the Exponential Rule which states that $\frac{d}{d y} [a^{y}]$ is $a^{y} \ln (a)$ where $a$ = $e$ .

$0 + e^{y}$

Step 4.2.1.1.4

Add $0$ and $e^{y}$ .

$e^{y}$

Step 4.2.1.2

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

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

Step 4.2.2

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

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

Step 4.2.3

Replace all occurrences of $u_{1}$ with $1 + e^{y}$ .

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

Step 4.3

Integrate the right side.

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Step 4.3.1

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

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

Step 4.3.2

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

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Step 4.3.2.1.1

Differentiate $1 + x^{2}$ .

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

Step 4.3.2.1.2

By the Sum Rule, the derivative of $1 + x^{2}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [x^{2}]$ .

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

Step 4.3.2.1.3

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

$0 + \frac{d}{d x} [x^{2}]$

Step 4.3.2.1.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$0 + 2 x$

Step 4.3.2.1.5

Add $0$ and $2 x$ .

$2 x$

Step 4.3.2.2

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

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

Step 4.3.3

Simplify.

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Step 4.3.3.1

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

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

Step 4.3.3.2

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

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

Step 4.3.4

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

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

Step 4.3.5

Simplify.

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Step 4.3.5.1

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

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

Step 4.3.5.2

Cancel the common factor of $2$ .

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Step 4.3.5.2.1

Cancel the common factor.

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

Step 4.3.5.2.2

Rewrite the expression.

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

Step 4.3.5.3

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

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

Step 4.3.6

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

$\ln (| 1 + e^{y} |) + C_{1} = \ln (| u_{2} |) + C_{2}$

Step 4.3.7

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

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

Step 4.4

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

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