Calculus Examples

$\frac{1}{x \sqrt{\ln (x) + 1}}$

Step 1

Write $\frac{1}{x \sqrt{\ln (x) + 1}}$ as a function.

$f (x) = \frac{1}{x \sqrt{\ln (x) + 1}}$

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.

$F (x) = \int \frac{1}{x \sqrt{\ln (x) + 1}} d x$

Step 4

Let $u = \ln (x) + 1$ . Then $d u = \frac{1}{x} d x$ , so $x d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.1

Let $u = \ln (x) + 1$ . Find $\frac{d u}{d x}$ .

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Step 4.1.1

Differentiate $\ln (x) + 1$ .

$\frac{d}{d x} [\ln (x) + 1]$

Step 4.1.2

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

$\frac{d}{d x} [\ln (x)] + \frac{d}{d x} [1]$

Step 4.1.3

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

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

Step 4.1.4

Differentiate using the Constant Rule.

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Step 4.1.4.1

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

$\frac{1}{x} + 0$

Step 4.1.4.2

Add $\frac{1}{x}$ and $0$ .

$\frac{1}{x}$

Step 4.2

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

$\int \frac{1}{\sqrt{u}} d u$

Step 5

Apply basic rules of exponents.

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Step 5.1

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

$\int \frac{1}{u^{\frac{1}{2}}} d u$

Step 5.2

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

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

Step 5.3

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

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Step 5.3.1

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

$\int u^{\frac{1}{2} \cdot - 1} d u$

Step 5.3.2

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

$\int u^{\frac{- 1}{2}} d u$

Step 5.3.3

Move the negative in front of the fraction.

$\int u^{- \frac{1}{2}} d u$

Step 6

By the Power Rule, the integral of $u^{- \frac{1}{2}}$ with respect to $u$ is $2 u^{\frac{1}{2}}$ .

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

Step 7

Replace all occurrences of $u$ with $\ln (x) + 1$ .

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

Step 8

The answer is the antiderivative of the function $f (x) = \frac{1}{x \sqrt{\ln (x) + 1}}$ .

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