Calculus Examples

$\int_{0}^{1} \ln (1 - x) d x$

Step 1

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \ln (1 - x)$ and $d v = 1$ .

$\ln (1 - x) x]_{0}^{1} - \int_{0}^{1} x (- \frac{1}{1 - x}) d x$

Step 2

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

$\ln (1 - x) x]_{0}^{1} - \int_{0}^{1} - \frac{x}{1 - x} d x$

Step 3

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

$\ln (1 - x) x]_{0}^{1} - - \int_{0}^{1} \frac{x}{1 - x} d x$

Step 4

Simplify the expression.

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Step 4.1

Simplify.

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Step 4.1.1

Multiply $- 1$ by $- 1$ .

$\ln (1 - x) x]_{0}^{1} + 1 \int_{0}^{1} \frac{x}{1 - x} d x$

Step 4.1.2

Multiply $\int_{0}^{1} \frac{x}{1 - x} d x$ by $1$ .

$\ln (1 - x) x]_{0}^{1} + \int_{0}^{1} \frac{x}{1 - x} d x$

Step 4.2

Reorder $1$ and $- x$ .

$\ln (1 - x) x]_{0}^{1} + \int_{0}^{1} \frac{x}{- x + 1} d x$

Step 5

Divide $x$ by $- x + 1$ .

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Step 5.1

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

$x$

$1$

$x$

$0$

Step 5.2

Divide the highest order term in the dividend $x$ by the highest order term in divisor $- x$ .

				-	$1$
-	$x$	+	$1$		$x$	+	$0$

Step 5.3

Multiply the new quotient term by the divisor.

				-	$1$
-	$x$	+	$1$		$x$	+	$0$
				+	$x$	-	$1$

Step 5.4

The expression needs to be subtracted from the dividend, so change all the signs in $x - 1$

				-	$1$
-	$x$	+	$1$		$x$	+	$0$
				-	$x$	+	$1$

Step 5.5

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

				-	$1$
-	$x$	+	$1$		$x$	+	$0$
				-	$x$	+	$1$
						+	$1$

Step 5.6

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

$\ln (1 - x) x]_{0}^{1} + \int_{0}^{1} - 1 + \frac{1}{- x + 1} d x$

Step 6

Split the single integral into multiple integrals.

$\ln (1 - x) x]_{0}^{1} + \int_{0}^{1} - 1 d x + \int_{0}^{1} \frac{1}{- x + 1} d x$

Step 7

Apply the constant rule.

$\ln (1 - x) x]_{0}^{1} + - x]_{0}^{1} + \int_{0}^{1} \frac{1}{- x + 1} d x$

Step 8

Let $u = - x + 1$ . Then $d u = - d x$ , so $- d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 8.1

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

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Step 8.1.1

Rewrite.

$\frac{- 1}{1}$

Step 8.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 8.2

Substitute the lower limit in for $x$ in $u = - x + 1$ .

$u_{lower} = - 0 + 1$

Step 8.3

Simplify.

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Step 8.3.1

Multiply $- 1$ by $0$ .

$u_{lower} = 0 + 1$

Step 8.3.2

Add $0$ and $1$ .

$u_{lower} = 1$

Step 8.4

Substitute the upper limit in for $x$ in $u = - x + 1$ .

$u_{upper} = - 1 \cdot 1 + 1$

Step 8.5

Simplify.

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Step 8.5.1

Multiply $- 1$ by $1$ .

$u_{upper} = - 1 + 1$

Step 8.5.2

Add $- 1$ and $1$ .

$u_{upper} = 0$

Step 8.6

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

$u_{lower} = 1$

$u_{upper} = 0$

Step 8.7

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$\ln (1 - x) x]_{0}^{1} + - x]_{0}^{1} + \int_{1}^{0} \frac{- 1}{u} d u$

Step 9

Move the negative in front of the fraction.

$\ln (1 - x) x]_{0}^{1} + - x]_{0}^{1} + \int_{1}^{0} - \frac{1}{u} d u$

Step 10

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

$\ln (1 - x) x]_{0}^{1} + - x]_{0}^{1} - \int_{1}^{0} \frac{1}{u} d u$

Step 11

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

$\ln (1 - x) x]_{0}^{1} + - x]_{0}^{1} - (\ln (| u |)]_{1}^{0})$

Step 12

Combine $\ln (1 - x) x]_{0}^{1}$ and $- x]_{0}^{1}$ .

$\ln (1 - x) x - x]_{0}^{1} - (\ln (| u |)]_{1}^{0})$

Step 13

Substitute and simplify.

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Step 13.1

Evaluate $\ln (1 - x) x - x$ at $1$ and at $0$ .

$(\ln (1 - 1 \cdot 1) \cdot 1 - 1 \cdot 1) - (\ln (1 - 0) \cdot 0 - 0) - (\ln (| u |)]_{1}^{0})$

Step 13.2

Evaluate $\ln (| u |)$ at $0$ and at $1$ .

$(\ln (1 - 1 \cdot 1) \cdot 1 - 1 \cdot 1) - (\ln (1 - 0) \cdot 0 - 0) - ((\ln (| 0 |)) - \ln (| 1 |))$

Step 13.3

Simplify.

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Step 13.3.1

Multiply $- 1$ by $1$ .

$\ln (1 - 1) \cdot 1 - 1 \cdot 1 - (\ln (1 - 0) \cdot 0 - 0) - ((\ln (| 0 |)) - \ln (| 1 |))$

Step 13.3.2

Subtract $1$ from $1$ .

$\ln (0) \cdot 1 - 1 \cdot 1 - (\ln (1 - 0) \cdot 0 - 0) - ((\ln (| 0 |)) - \ln (| 1 |))$

Step 13.3.3

The natural logarithm of zero is undefined.

Undefined

$\ln (0) \cdot 1 - 1 \cdot 1 - (\ln (1 - 0) \cdot 0 - 0) - ((\ln (| 0 |)) - \ln (| 1 |))$

Step 13.4

The natural logarithm of zero is undefined.

Undefined

$\ln (0) \cdot 1 - 1 \cdot 1 - (\ln (1 - 0) \cdot 0 - 0) - ((\ln (| 0 |)) - \ln (| 1 |))$

Step 14

The natural logarithm of zero is undefined.

Undefined