Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Solve for $\frac{d f}{d x}$ .

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Step 1.1.1

Reorder factors in $e^{2 x} \frac{d f}{d x} + e^{x}$ .

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

Step 1.1.2

Subtract $e^{x}$ from both sides of the equation.

$\frac{d f}{d x} e^{2 x} = 1 - e^{x}$

Step 1.1.3

Divide each term in $\frac{d f}{d x} e^{2 x} = 1 - e^{x}$ by $e^{2 x}$ and simplify.

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Step 1.1.3.1

Divide each term in $\frac{d f}{d x} e^{2 x} = 1 - e^{x}$ by $e^{2 x}$ .

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

Step 1.1.3.2

Simplify the left side.

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Step 1.1.3.2.1

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

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Step 1.1.3.2.1.1

Cancel the common factor.

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

Step 1.1.3.2.1.2

Divide $\frac{d f}{d x}$ by $1$ .

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

Step 1.1.3.3

Simplify the right side.

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Step 1.1.3.3.1

Cancel the common factor of $e^{x}$ and $e^{2 x}$ .

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Step 1.1.3.3.1.1

Factor $e^{2 x}$ out of $- e^{x}$ .

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

Step 1.1.3.3.1.2

Cancel the common factors.

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Step 1.1.3.3.1.2.1

Multiply by $1$ .

$\frac{d f}{d x} = \frac{1}{e^{2 x}} + \frac{e^{2 x} (- e^{- x})}{e^{2 x} \cdot 1}$

Step 1.1.3.3.1.2.2

Cancel the common factor.

$\frac{d f}{d x} = \frac{1}{e^{2 x}} + \frac{e^{2 x} (- e^{- x})}{e^{2 x} \cdot 1}$

Step 1.1.3.3.1.2.3

Rewrite the expression.

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

Step 1.1.3.3.1.2.4

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

$\frac{d f}{d x} = \frac{1}{e^{2 x}} - e^{- x}$

Step 1.2

Rewrite the equation.

$d f = (\frac{1}{e^{2 x}} - e^{- x}) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d f = \int \frac{1}{e^{2 x}} - e^{- x} d x$

Step 2.2

Apply the constant rule.

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

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

Step 2.3.2

Simplify the expression.

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Step 2.3.2.1

Negate the exponent of $e^{2 x}$ and move it out of the denominator.

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

Step 2.3.2.2

Simplify.

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Step 2.3.2.2.1

Multiply the exponents in ${(e^{2 x})}^{- 1}$ .

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Step 2.3.2.2.1.1

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

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

Step 2.3.2.2.1.2

Multiply $- 1$ by $2$ .

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

Step 2.3.2.2.2

Multiply $e^{- 2 x}$ by $1$ .

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

Step 2.3.3

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

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Step 2.3.3.1

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

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Step 2.3.3.1.1

Differentiate $- 2 x$ .

$\frac{d}{d x} [- 2 x]$

Step 2.3.3.1.2

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x$ with respect to $x$ is $- 2 \frac{d}{d x} [x]$ .

$- 2 \frac{d}{d x} [x]$

Step 2.3.3.1.3

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

$- 2 \cdot 1$

Step 2.3.3.1.4

Multiply $- 2$ by $1$ .

$- 2$

Step 2.3.3.2

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

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

Step 2.3.4

Simplify.

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Step 2.3.4.1

Move the negative in front of the fraction.

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

Step 2.3.4.2

Combine $e^{u_{1}}$ and $\frac{1}{2}$ .

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

Step 2.3.5

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

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

Step 2.3.6

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

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

Step 2.3.7

The integral of $e^{u_{1}}$ with respect to $u_{1}$ is $e^{u_{1}}$ .

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

Step 2.3.8

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

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

Step 2.3.9

Let $u_{2} = - x$ . Then $d u_{2} = - d x$ , so $- d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 2.3.9.1

Let $u_{2} = - x$ . Find $\frac{d u_{2}}{d x}$ .

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Step 2.3.9.1.1

Differentiate $- x$ .

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

Step 2.3.9.1.2

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 2.3.9.1.3

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 2.3.9.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 2.3.9.2

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

$f + C_{1} = - \frac{1}{2} (e^{u_{1}} + C_{2}) - \int - e^{u_{2}} d u_{2}$

Step 2.3.10

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

$f + C_{1} = - \frac{1}{2} (e^{u_{1}} + C_{2}) - - \int e^{u_{2}} d u_{2}$

Step 2.3.11

Simplify.

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Step 2.3.11.1

Multiply $- 1$ by $- 1$ .

$f + C_{1} = - \frac{1}{2} (e^{u_{1}} + C_{2}) + 1 \int e^{u_{2}} d u_{2}$

Step 2.3.11.2

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

$f + C_{1} = - \frac{1}{2} (e^{u_{1}} + C_{2}) + \int e^{u_{2}} d u_{2}$

Step 2.3.12

The integral of $e^{u_{2}}$ with respect to $u_{2}$ is $e^{u_{2}}$ .

$f + C_{1} = - \frac{1}{2} (e^{u_{1}} + C_{2}) + e^{u_{2}} + C_{3}$

Step 2.3.13

Simplify.

$f + C_{1} = - \frac{1}{2} e^{u_{1}} + e^{u_{2}} + C_{4}$

Step 2.3.14

Substitute back in for each integration substitution variable.

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Step 2.3.14.1

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

$f + C_{1} = - \frac{1}{2} e^{- 2 x} + e^{u_{2}} + C_{4}$

Step 2.3.14.2

Replace all occurrences of $u_{2}$ with $- x$ .

$f + C_{1} = - \frac{1}{2} e^{- 2 x} + e^{- x} + C_{4}$

Step 2.4

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

$f = - \frac{1}{2} e^{- 2 x} + e^{- x} + K$