Calculus Examples

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

Step 1

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

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Step 1.1

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

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Step 1.1.1

Differentiate $x + 1$ .

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

Step 1.1.2

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

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

Step 1.1.3

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

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

Step 1.1.4

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

$1 + 0$

Step 1.1.5

Add $1$ and $0$ .

$1$

Step 1.2

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

$\int \ln (\frac{\sqrt{1 - (u_{1} - 1)}}{u_{1}}) d u_{1}$

Step 2

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \ln (\frac{\sqrt{1 - (u_{1} - 1)}}{u_{1}})$ and $d v = 1$ .

$\ln (\frac{\sqrt{1 - (u_{1} - 1)}}{u_{1}}) u_{1} - \int u_{1} (- \frac{u_{1} - 4}{(2 u_{1}) (u_{1} - 2)}) d u_{1}$

Step 3

Simplify.

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Step 3.1

Combine $u_{1}$ and $\frac{u_{1} - 4}{2 u_{1} (u_{1} - 2)}$ .

$\ln (\frac{\sqrt{1 - (u_{1} - 1)}}{u_{1}}) u_{1} - \int - \frac{u_{1} (u_{1} - 4)}{2 u_{1} (u_{1} - 2)} d u_{1}$

Step 3.2

Cancel the common factor of $u_{1}$ .

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Step 3.2.1

Cancel the common factor.

$\ln (\frac{\sqrt{1 - (u_{1} - 1)}}{u_{1}}) u_{1} - \int - \frac{u_{1} (u_{1} - 4)}{2 u_{1} (u_{1} - 2)} d u_{1}$

Step 3.2.2

Rewrite the expression.

$\ln (\frac{\sqrt{1 - (u_{1} - 1)}}{u_{1}}) u_{1} - \int - \frac{u_{1} - 4}{2 (u_{1} - 2)} d u_{1}$

Step 4

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

$\ln (\frac{\sqrt{1 - (u_{1} - 1)}}{u_{1}}) u_{1} - - \int \frac{u_{1} - 4}{2 (u_{1} - 2)} d u_{1}$

Step 5

Simplify.

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Step 5.1

Multiply $- 1$ by $- 1$ .

$\ln (\frac{\sqrt{1 - (u_{1} - 1)}}{u_{1}}) u_{1} + 1 \int \frac{u_{1} - 4}{2 (u_{1} - 2)} d u_{1}$

Step 5.2

Multiply $\int \frac{u_{1} - 4}{2 (u_{1} - 2)} d u_{1}$ by $1$ .

$\ln (\frac{\sqrt{1 - (u_{1} - 1)}}{u_{1}}) u_{1} + \int \frac{u_{1} - 4}{2 (u_{1} - 2)} d u_{1}$

Step 6

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

$\ln (\frac{\sqrt{1 - (u_{1} - 1)}}{u_{1}}) u_{1} + \frac{1}{2} \int \frac{u_{1} - 4}{u_{1} - 2} d u_{1}$

Step 7

Divide $u_{1} - 4$ by $u_{1} - 2$ .

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Step 7.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$u_{1}$

$2$

$u_{1}$

$4$

Step 7.2

Divide the highest order term in the dividend $u_{1}$ by the highest order term in divisor $u_{1}$ .

					$1$
	$u_{1}$	-	$2$		$u_{1}$	-	$4$

Step 7.3

Multiply the new quotient term by the divisor.

				$1$
$u_{1}$	-	$2$		$u_{1}$	-	$4$
			+	$u_{1}$	-	$2$

Step 7.4

The expression needs to be subtracted from the dividend, so change all the signs in $u_{1} - 2$

				$1$
$u_{1}$	-	$2$		$u_{1}$	-	$4$
			-	$u_{1}$	+	$2$

Step 7.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$1$
$u_{1}$	-	$2$		$u_{1}$	-	$4$
			-	$u_{1}$	+	$2$
					-	$2$

Step 7.6

The final answer is the quotient plus the remainder over the divisor.

$\ln (\frac{\sqrt{1 - (u_{1} - 1)}}{u_{1}}) u_{1} + \frac{1}{2} \int 1 - \frac{2}{u_{1} - 2} d u_{1}$

Step 8

Split the single integral into multiple integrals.

$\ln (\frac{\sqrt{1 - (u_{1} - 1)}}{u_{1}}) u_{1} + \frac{1}{2} (\int d u_{1} + \int - \frac{2}{u_{1} - 2} d u_{1})$

Step 9

Apply the constant rule.

$\ln (\frac{\sqrt{1 - (u_{1} - 1)}}{u_{1}}) u_{1} + \frac{1}{2} (u_{1} + C + \int - \frac{2}{u_{1} - 2} d u_{1})$

Step 10

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

$\ln (\frac{\sqrt{1 - (u_{1} - 1)}}{u_{1}}) u_{1} + \frac{1}{2} (u_{1} + C - \int \frac{2}{u_{1} - 2} d u_{1})$

Step 11

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

$\ln (\frac{\sqrt{1 - (u_{1} - 1)}}{u_{1}}) u_{1} + \frac{1}{2} (u_{1} + C - (2 \int \frac{1}{u_{1} - 2} d u_{1}))$

Step 12

Multiply $2$ by $- 1$ .

$\ln (\frac{\sqrt{1 - (u_{1} - 1)}}{u_{1}}) u_{1} + \frac{1}{2} (u_{1} + C - 2 \int \frac{1}{u_{1} - 2} d u_{1})$

Step 13

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

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Step 13.1

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

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Step 13.1.1

Differentiate $u_{1} - 2$ .

$\frac{d}{d u_{1}} [u_{1} - 2]$

Step 13.1.2

By the Sum Rule, the derivative of $u_{1} - 2$ with respect to $u_{1}$ is $\frac{d}{d u_{1}} [u_{1}] + \frac{d}{d u_{1}} [- 2]$ .

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

Step 13.1.3

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

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

Step 13.1.4

Since $- 2$ is constant with respect to $u_{1}$ , the derivative of $- 2$ with respect to $u_{1}$ is $0$ .

$1 + 0$

Step 13.1.5

Add $1$ and $0$ .

$1$

Step 13.2

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

$\ln (\frac{\sqrt{1 - (u_{1} - 1)}}{u_{1}}) u_{1} + \frac{1}{2} (u_{1} + C - 2 \int \frac{1}{u_{2}} d u_{2})$

Step 14

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$\ln (\frac{\sqrt{1 - (u_{1} - 1)}}{u_{1}}) u_{1} + \frac{1}{2} (u_{1} + C - 2 (\ln (| u_{2} |) + C))$

Step 15

Simplify.

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Step 15.1

Simplify.

$\ln (\frac{\sqrt{- u_{1} + 2}}{u_{1}}) u_{1} + \frac{1}{2} (u_{1} - 2 \ln (| u_{2} |)) + C$

Step 15.2

Rewrite $\ln (\frac{\sqrt{- u_{1} + 2}}{u_{1}}) u_{1} + \frac{1}{2} (u_{1} - 2 \ln (| u_{2} |)) + C$ as $\ln (\frac{\sqrt{- u_{1} + 2}}{u_{1}}) u_{1} + \frac{u_{1}}{2} - \ln (| u_{2} |) + C$ .

$\ln (\frac{\sqrt{- u_{1} + 2}}{u_{1}}) u_{1} + \frac{u_{1}}{2} - \ln (| u_{2} |) + C$

Step 15.3

Simplify.

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Step 15.3.1

To write $\ln (\frac{\sqrt{- u_{1} + 2}}{u_{1}}) u_{1}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\ln (\frac{\sqrt{- u_{1} + 2}}{u_{1}}) u_{1} \cdot \frac{2}{2} + \frac{u_{1}}{2} - \ln (| u_{2} |) + C$

Step 15.3.2

Combine $\ln (\frac{\sqrt{- u_{1} + 2}}{u_{1}}) u_{1}$ and $\frac{2}{2}$ .

$\frac{\ln (\frac{\sqrt{- u_{1} + 2}}{u_{1}}) u_{1} \cdot 2}{2} + \frac{u_{1}}{2} - \ln (| u_{2} |) + C$

Step 15.3.3

Combine the numerators over the common denominator.

$\frac{\ln (\frac{\sqrt{- u_{1} + 2}}{u_{1}}) u_{1} \cdot 2 + u_{1}}{2} - \ln (| u_{2} |) + C$

Step 15.3.4

Move $2$ to the left of $\ln (\frac{\sqrt{- u_{1} + 2}}{u_{1}}) u_{1}$ .

$\frac{2 \ln (\frac{\sqrt{- u_{1} + 2}}{u_{1}}) u_{1} + u_{1}}{2} - \ln (| u_{2} |) + C$

Step 16

Substitute back in for each integration substitution variable.

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Step 16.1

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

$\frac{2 \ln (\frac{\sqrt{- (x + 1) + 2}}{x + 1}) (x + 1) + x + 1}{2} - \ln (| u_{2} |) + C$

Step 16.2

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

$\frac{2 \ln (\frac{\sqrt{- (x + 1) + 2}}{x + 1}) (x + 1) + x + 1}{2} - \ln (| u_{1} - 2 |) + C$

Step 16.3

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

$\frac{2 \ln (\frac{\sqrt{- (x + 1) + 2}}{x + 1}) (x + 1) + x + 1}{2} - \ln (| x + 1 - 2 |) + C$

Step 17

Simplify.

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Step 17.1

Apply the distributive property.

$\frac{2 \ln (\frac{\sqrt{- x - 1 \cdot 1 + 2}}{x + 1}) (x + 1) + x + 1}{2} - \ln (| x + 1 - 2 |) + C$

Step 17.2

Multiply $- 1$ by $1$ .

$\frac{2 \ln (\frac{\sqrt{- x - 1 + 2}}{x + 1}) (x + 1) + x + 1}{2} - \ln (| x + 1 - 2 |) + C$

Step 17.3

Add $- 1$ and $2$ .

$\frac{2 \ln (\frac{\sqrt{- x + 1}}{x + 1}) (x + 1) + x + 1}{2} - \ln (| x + 1 - 2 |) + C$

Step 17.4

Subtract $2$ from $1$ .

$\frac{2 \ln (\frac{\sqrt{- x + 1}}{x + 1}) (x + 1) + x + 1}{2} - \ln (| x - 1 |) + C$

Step 18

Reorder terms.

$\frac{1}{2} (2 \ln (\frac{\sqrt{- x + 1}}{x + 1}) (x + 1) + x + 1) - \ln (| x - 1 |) + C$