Calculus Examples

$\int_{0}^{1} \frac{e^{2 x} - e^{- 2 x}}{e^{2 x} + e^{- 2 x}} d x$

Step 1

Let $u_{3} = e^{2 x} + e^{- 2 x}$ . Then $d u_{3} = (2 e^{2 x} - 2 e^{- 2 x}) d x$ , so $\frac{1}{2} d u_{3} = (e^{2 x} - e^{- 2 x}) d x$ . Rewrite using $u_{3}$ and $d$ $u_{3}$ .

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Step 1.1

Let $u_{3} = e^{2 x} + e^{- 2 x}$ . Find $\frac{d u_{3}}{d x}$ .

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Step 1.1.1

Differentiate $e^{2 x} + e^{- 2 x}$ .

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

Step 1.1.2

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

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

Step 1.1.3

Evaluate $\frac{d}{d x} [e^{2 x}]$ .

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Step 1.1.3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 2 x$ .

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Step 1.1.3.1.1

To apply the Chain Rule, set $u_{1}$ as $2 x$ .

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

Step 1.1.3.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

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

Step 1.1.3.1.3

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

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

Step 1.1.3.2

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

$e^{2 x} (2 \frac{d}{d x} [x]) + \frac{d}{d x} [e^{- 2 x}]$

Step 1.1.3.3

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

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

Step 1.1.3.4

Multiply $2$ by $1$ .

$e^{2 x} \cdot 2 + \frac{d}{d x} [e^{- 2 x}]$

Step 1.1.3.5

Move $2$ to the left of $e^{2 x}$ .

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

Step 1.1.4

Evaluate $\frac{d}{d x} [e^{- 2 x}]$ .

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Step 1.1.4.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - 2 x$ .

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Step 1.1.4.1.1

To apply the Chain Rule, set $u_{2}$ as $- 2 x$ .

$2 e^{2 x} + \frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [- 2 x]$

Step 1.1.4.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$2 e^{2 x} + e^{u_{2}} \frac{d}{d x} [- 2 x]$

Step 1.1.4.1.3

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

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

Step 1.1.4.2

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

$2 e^{2 x} + e^{- 2 x} (- 2 \frac{d}{d x} [x])$

Step 1.1.4.3

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

$2 e^{2 x} + e^{- 2 x} (- 2 \cdot 1)$

Step 1.1.4.4

Multiply $- 2$ by $1$ .

$2 e^{2 x} + e^{- 2 x} \cdot - 2$

Step 1.1.4.5

Move $- 2$ to the left of $e^{- 2 x}$ .

$2 e^{2 x} - 2 e^{- 2 x}$

Step 1.2

Substitute the lower limit in for $x$ in $u_{3} = e^{2 x} + e^{- 2 x}$ .

$u_{lower} = e^{2 \cdot 0} + e^{- 2 \cdot 0}$

Step 1.3

Simplify.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

Multiply $2$ by $0$ .

$u_{lower} = e^{0} + e^{- 2 \cdot 0}$

Step 1.3.1.2

Anything raised to $0$ is $1$ .

$u_{lower} = 1 + e^{- 2 \cdot 0}$

Step 1.3.1.3

Multiply $- 2$ by $0$ .

$u_{lower} = 1 + e^{0}$

Step 1.3.1.4

Anything raised to $0$ is $1$ .

$u_{lower} = 1 + 1$

Step 1.3.2

Add $1$ and $1$ .

$u_{lower} = 2$

Step 1.4

Substitute the upper limit in for $x$ in $u_{3} = e^{2 x} + e^{- 2 x}$ .

$u_{upper} = e^{2 \cdot 1} + e^{- 2 \cdot 1}$

Step 1.5

Simplify each term.

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Step 1.5.1

Multiply $2$ by $1$ .

$u_{upper} = e^{2} + e^{- 2 \cdot 1}$

Step 1.5.2

Multiply $- 2$ by $1$ .

$u_{upper} = e^{2} + e^{- 2}$

Step 1.5.3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$u_{upper} = e^{2} + \frac{1}{e^{2}}$

Step 1.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 2$

$u_{upper} = e^{2} + \frac{1}{e^{2}}$

Step 1.7

Rewrite the problem using $u_{3}$ , $d u_{3}$ , and the new limits of integration.

$\int_{2}^{e^{2} + \frac{1}{e^{2}}} \frac{1}{u_{3}} \cdot \frac{1}{2} d u_{3}$

Step 2

Simplify.

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Step 2.1

Multiply $\frac{1}{u_{3}}$ by $\frac{1}{2}$ .

$\int_{2}^{e^{2} + \frac{1}{e^{2}}} \frac{1}{u_{3} \cdot 2} d u_{3}$

Step 2.2

Move $2$ to the left of $u_{3}$ .

$\int_{2}^{e^{2} + \frac{1}{e^{2}}} \frac{1}{2 u_{3}} d u_{3}$

Step 3

Since $\frac{1}{2}$ is constant with respect to $u_{3}$ , move $\frac{1}{2}$ out of the integral.

$\frac{1}{2} \int_{2}^{e^{2} + \frac{1}{e^{2}}} \frac{1}{u_{3}} d u_{3}$

Step 4

The integral of $\frac{1}{u_{3}}$ with respect to $u_{3}$ is $\ln (| u_{3} |)$ .

$\frac{1}{2} \ln (| u_{3} |)]_{2}^{e^{2} + \frac{1}{e^{2}}}$

Step 5

Evaluate $\ln (| u_{3} |)$ at $e^{2} + \frac{1}{e^{2}}$ and at $2$ .

$\frac{1}{2} (\ln (| e^{2} + \frac{1}{e^{2}} |) - \ln (| 2 |))$

Step 6

Simplify.

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Step 6.1

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\frac{1}{2} \ln (\frac{| e^{2} + \frac{1}{e^{2}} |}{| 2 |})$

Step 6.2

Combine $\frac{1}{2}$ and $\ln (\frac{| e^{2} + \frac{1}{e^{2}} |}{| 2 |})$ .

$\frac{\ln (\frac{| e^{2} + \frac{1}{e^{2}} |}{| 2 |})}{2}$

Step 7

Simplify.

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Step 7.1

Simplify the numerator.

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Step 7.1.1

$e^{2} + \frac{1}{e^{2}}$ is approximately $7.52439138$ which is positive so remove the absolute value

$\frac{\ln (\frac{e^{2} + \frac{1}{e^{2}}}{| 2 |})}{2}$

Step 7.1.2

To write $e^{2}$ as a fraction with a common denominator, multiply by $\frac{e^{2}}{e^{2}}$ .

$\frac{\ln (\frac{\frac{e^{2} e^{2}}{e^{2}} + \frac{1}{e^{2}}}{| 2 |})}{2}$

Step 7.1.3

Combine the numerators over the common denominator.

$\frac{\ln (\frac{\frac{e^{2} e^{2} + 1}{e^{2}}}{| 2 |})}{2}$

Step 7.1.4

Multiply $e^{2}$ by $e^{2}$ by adding the exponents.

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Step 7.1.4.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{\ln (\frac{\frac{e^{2 + 2} + 1}{e^{2}}}{| 2 |})}{2}$

Step 7.1.4.2

Add $2$ and $2$ .

$\frac{\ln (\frac{\frac{e^{4} + 1}{e^{2}}}{| 2 |})}{2}$

Step 7.2

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{\ln (\frac{\frac{e^{4} + 1}{e^{2}}}{2})}{2}$

Step 7.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{\ln (\frac{e^{4} + 1}{e^{2}} \cdot \frac{1}{2})}{2}$

Step 7.4

Multiply $\frac{e^{4} + 1}{e^{2}}$ by $\frac{1}{2}$ .

$\frac{\ln (\frac{e^{4} + 1}{e^{2} \cdot 2})}{2}$

Step 7.5

Move $2$ to the left of $e^{2}$ .

$\frac{\ln (\frac{e^{4} + 1}{2 e^{2}})}{2}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$\frac{\ln (\frac{e^{4} + 1}{2 e^{2}})}{2}$

Decimal Form:

$0.66250137 \dots$

Step 9