Calculus Examples

$2^{\sqrt{x}} \cdot \frac{\ln (2)}{\sqrt{x}}$

Step 1

Write $2^{\sqrt{x}} \cdot \frac{\ln (2)}{\sqrt{x}}$ as a function.

$f (x) = 2^{\sqrt{x}} \cdot \frac{\ln (2)}{\sqrt{x}}$

Step 2

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Step 3

Set up the integral to solve.

$F (x) = \int 2^{\sqrt{x}} \cdot \frac{\ln (2)}{\sqrt{x}} d x$

Step 4

Combine $2^{\sqrt{x}}$ and $\frac{\ln (2)}{\sqrt{x}}$ .

$\int \frac{2^{\sqrt{x}} \ln (2)}{\sqrt{x}} d x$

Step 5

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

$\ln (2) \int \frac{2^{\sqrt{x}}}{\sqrt{x}} d x$

Step 6

Apply basic rules of exponents.

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Step 6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

$\ln (2) \int \frac{2^{x^{\frac{1}{2}}}}{\sqrt{x}} d x$

Step 6.2

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

$\ln (2) \int \frac{2^{x^{\frac{1}{2}}}}{x^{\frac{1}{2}}} d x$

Step 6.3

Move $x^{\frac{1}{2}}$ out of the denominator by raising it to the $- 1$ power.

$\ln (2) \int 2^{x^{\frac{1}{2}}} {(x^{\frac{1}{2}})}^{- 1} d x$

Step 6.4

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

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Step 6.4.1

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

$\ln (2) \int 2^{x^{\frac{1}{2}}} x^{\frac{1}{2} \cdot - 1} d x$

Step 6.4.2

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

$\ln (2) \int 2^{x^{\frac{1}{2}}} x^{\frac{- 1}{2}} d x$

Step 6.4.3

Move the negative in front of the fraction.

$\ln (2) \int 2^{x^{\frac{1}{2}}} x^{- \frac{1}{2}} d x$

Step 7

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

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Step 7.1

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

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Step 7.1.1

Differentiate $x^{\frac{1}{2}}$ .

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

Step 7.1.2

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

$\frac{1}{2} x^{\frac{1}{2} - 1}$

Step 7.1.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{1}{2} x^{\frac{1}{2} - 1 \cdot \frac{2}{2}}$

Step 7.1.4

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

$\frac{1}{2} x^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}}$

Step 7.1.5

Combine the numerators over the common denominator.

$\frac{1}{2} x^{\frac{1 - 1 \cdot 2}{2}}$

Step 7.1.6

Simplify the numerator.

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Step 7.1.6.1

Multiply $- 1$ by $2$ .

$\frac{1}{2} x^{\frac{1 - 2}{2}}$

Step 7.1.6.2

Subtract $2$ from $1$ .

$\frac{1}{2} x^{\frac{- 1}{2}}$

Step 7.1.7

Move the negative in front of the fraction.

$\frac{1}{2} x^{- \frac{1}{2}}$

Step 7.1.8

Simplify.

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Step 7.1.8.1

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

$\frac{1}{2} \cdot \frac{1}{x^{\frac{1}{2}}}$

Step 7.1.8.2

Multiply $\frac{1}{2}$ by $\frac{1}{x^{\frac{1}{2}}}$ .

$\frac{1}{2 x^{\frac{1}{2}}}$

Step 7.2

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

$\ln (2) \int 2 \cdot 2^{u_{1}} d u_{1}$

Step 8

Multiply $2$ by $2^{u_{1}}$ .

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Step 8.1

Raise $2$ to the power of $1$ .

$\ln (2) \int 2^{1} \cdot 2^{u_{1}} d u_{1}$

Step 8.2

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

$\ln (2) \int 2^{1 + u_{1}} d u_{1}$

Step 9

Let $u_{2} = 1 + u_{1}$ . Then $d u_{2} = d u_{1}$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 9.1

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

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Step 9.1.1

Differentiate $1 + u_{1}$ .

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

Step 9.1.2

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

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

Step 9.1.3

Since $1$ is constant with respect to $u_{1}$ , the derivative of $1$ with respect to $u_{1}$ is $0$ .

$0 + \frac{d}{d u_{1}} [u_{1}]$

Step 9.1.4

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

$0 + 1$

Step 9.1.5

Add $0$ and $1$ .

$1$

Step 9.2

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

$\ln (2) \int 2^{u_{2}} d u_{2}$

Step 10

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

$\ln (2) (\frac{2^{u_{2}}}{\ln (2)} + C)$

Step 11

Simplify.

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Step 11.1

Rewrite $\ln (2) (\frac{2^{u_{2}}}{\ln (2)} + C)$ as $\ln (2) \frac{1}{\ln (2)} \cdot 2^{u_{2}} + C$ .

$\ln (2) \frac{1}{\ln (2)} \cdot 2^{u_{2}} + C$

Step 11.2

Simplify.

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Step 11.2.1

Combine $\ln (2)$ and $\frac{1}{\ln (2)}$ .

$\frac{\ln (2)}{\ln (2)} \cdot 2^{u_{2}} + C$

Step 11.2.2

Cancel the common factor of $\ln (2)$ .

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Step 11.2.2.1

Cancel the common factor.

$\frac{\ln (2)}{\ln (2)} \cdot 2^{u_{2}} + C$

Step 11.2.2.2

Rewrite the expression.

$1 \cdot 2^{u_{2}} + C$

Step 11.2.3

Multiply $2^{u_{2}}$ by $1$ .

$2^{u_{2}} + C$

Step 12

Substitute back in for each integration substitution variable.

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Step 12.1

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

$2^{1 + u_{1}} + C$

Step 12.2

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

$2^{1 + x^{\frac{1}{2}}} + C$

Step 13

The answer is the antiderivative of the function $f (x) = 2^{\sqrt{x}} \cdot \frac{\ln (2)}{\sqrt{x}}$ .

$F (x) =$ $2^{1 + x^{\frac{1}{2}}} + C$