Calculus Examples

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

Step 1

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

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

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

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Step 3.1.1

Cancel the common factor.

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

Step 3.1.2

Rewrite the expression.

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

Step 3.2

Rewrite using the commutative property of multiplication.

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

Step 3.3

Apply the distributive property.

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

Step 3.4

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

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Step 3.4.1

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

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

Step 3.4.2

Cancel the common factor.

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

Step 3.4.3

Rewrite the expression.

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

Step 3.5

Multiply $- 1$ by $1$ .

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Apply the constant rule.

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

Step 4.3

Integrate the right side.

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Step 4.3.1

Split the single integral into multiple integrals.

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

Step 4.3.2

Apply the constant rule.

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

Step 4.3.3

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

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

Step 4.3.4

Simplify the expression.

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Step 4.3.4.1

Negate the exponent of $e^{x}$ and move it out of the denominator.

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

Step 4.3.4.2

Simplify.

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Step 4.3.4.2.1

Multiply the exponents in ${(e^{x})}^{- 1}$ .

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Step 4.3.4.2.1.1

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

$y + C_{1} = - x + C_{2} - \int 1 e^{x \cdot - 1} d x$

Step 4.3.4.2.1.2

Move $- 1$ to the left of $x$ .

$y + C_{1} = - x + C_{2} - \int 1 e^{- 1 \cdot x} d x$

Step 4.3.4.2.1.3

Rewrite $- 1 x$ as $- x$ .

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

Step 4.3.4.2.2

Multiply $e^{- x}$ by $1$ .

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

Step 4.3.5

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

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Step 4.3.5.1

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

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Step 4.3.5.1.1

Differentiate $- x$ .

$\frac{d}{d x} [- x]$

Step 4.3.5.1.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$- \frac{d}{d x} [x]$

Step 4.3.5.1.3

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

$- 1 \cdot 1$

Step 4.3.5.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 4.3.5.2

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

$y + C_{1} = - x + C_{2} - \int - e^{u} d u$

Step 4.3.6

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

$y + C_{1} = - x + C_{2} - - \int e^{u} d u$

Step 4.3.7

Simplify.

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Step 4.3.7.1

Multiply $- 1$ by $- 1$ .

$y + C_{1} = - x + C_{2} + 1 \int e^{u} d u$

Step 4.3.7.2

Multiply $\int e^{u} d u$ by $1$ .

$y + C_{1} = - x + C_{2} + \int e^{u} d u$

Step 4.3.8

The integral of $e^{u}$ with respect to $u$ is $e^{u}$ .

$y + C_{1} = - x + C_{2} + e^{u} + C_{3}$

Step 4.3.9

Simplify.

$y + C_{1} = - x + e^{u} + C_{4}$

Step 4.3.10

Replace all occurrences of $u$ with $- x$ .

$y + C_{1} = - x + e^{- x} + C_{4}$

Step 4.4

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

$y = - x + e^{- x} + K$