Calculus Examples

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

Step 1

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

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

Rewrite using the commutative property of multiplication.

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

Cancel the common factor.

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

Step 3.2.3

Rewrite the expression.

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

Step 3.3

Move the negative in front of the fraction.

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

Step 3.4

Rewrite using the commutative property of multiplication.

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

Step 3.5

Cancel the common factor of $y + 1$ .

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Step 3.5.1

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

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

Step 3.5.2

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

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

Step 3.5.3

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

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

Step 3.5.4

Cancel the common factor.

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

Step 3.5.5

Rewrite the expression.

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

Step 3.6

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

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

Step 3.7

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

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.

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

Step 4.2.2

Let $u_{1} = y + 1$ . Then $d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 4.2.2.1

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

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Step 4.2.2.1.1

Differentiate $y + 1$ .

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

Step 4.2.2.1.2

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

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

Step 4.2.2.1.3

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

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

Step 4.2.2.1.4

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

$1 + 0$

Step 4.2.2.1.5

Add $1$ and $0$ .

$1$

Step 4.2.2.2

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

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

Step 4.2.3

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

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

Step 4.2.4

Simplify.

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

Step 4.2.5

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

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

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.

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

Step 4.3.2

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

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Step 4.3.2.1

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

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Step 4.3.2.1.1

Differentiate $e^{x} + 1$ .

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

Step 4.3.2.1.2

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

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

Step 4.3.2.1.3

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

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

Step 4.3.2.1.4

Differentiate using the Constant Rule.

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Step 4.3.2.1.4.1

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

$e^{x} + 0$

Step 4.3.2.1.4.2

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

$e^{x}$

Step 4.3.2.2

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

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

Step 4.3.3

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

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

Step 4.3.4

Simplify.

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

Step 4.3.5

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

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

Step 4.4

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

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