Calculus Examples

$\int \ln (1 - x) d x$

Step 1

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

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

Step 2

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

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

Step 3

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

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

Step 4

Simplify the expression.

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Step 4.1

Simplify.

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Step 4.1.1

Multiply $- 1$ by $- 1$ .

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

Step 4.1.2

Multiply $\int \frac{x}{1 - x} d x$ by $1$ .

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

Step 4.2

Reorder $1$ and $- x$ .

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

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.

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

Step 6

Split the single integral into multiple integrals.

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

Step 7

Apply the constant rule.

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

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

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

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

Step 9

Move the negative in front of the fraction.

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

Step 10

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

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

Step 11

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

$\ln (1 - x) x - x + C - (\ln (| u |) + C)$

Step 12

Simplify.

$\ln (1 - x) x - x - \ln (| u |) + C$

Step 13

Replace all occurrences of $u$ with $- x + 1$ .

$\ln (1 - x) x - x - \ln (| - x + 1 |) + C$