Calculus Examples

$\int \frac{x - 5}{- 2 x + 2} d x$

Step 1

Divide $x - 5$ by $- 2 x + 2$ .

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Step 1.1

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

$2 x$

$2$

$x$

$5$

Step 1.2

Divide the highest order term in the dividend $x$ by the highest order term in divisor $- 2 x$ .

				-	$\frac{1}{2}$
-	$2 x$	+	$2$		$x$	-	$5$

Step 1.3

Multiply the new quotient term by the divisor.

				-	$\frac{1}{2}$
-	$2 x$	+	$2$		$x$	-	$5$
				+	$x$	-	$1$

Step 1.4

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

				-	$\frac{1}{2}$
-	$2 x$	+	$2$		$x$	-	$5$
				-	$x$	+	$1$

Step 1.5

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

				-	$\frac{1}{2}$
-	$2 x$	+	$2$		$x$	-	$5$
				-	$x$	+	$1$
						-	$4$

Step 1.6

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

$\int - \frac{1}{2} - \frac{4}{- 2 x + 2} d x$

Step 2

Split the single integral into multiple integrals.

$\int - \frac{1}{2} d x + \int - \frac{4}{- 2 x + 2} d x$

Step 3

Cancel the common factor of $4$ and $- 2 x + 2$ .

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Step 3.1

Factor $2$ out of $4$ .

$\int - \frac{1}{2} d x + \int - \frac{2 \cdot 2}{- 2 x + 2} d x$

Step 3.2

Cancel the common factors.

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Step 3.2.1

Factor $2$ out of $- 2 x$ .

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

Step 3.2.2

Factor $2$ out of $2$ .

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

Step 3.2.3

Factor $2$ out of $2 (- x) + 2 (1)$ .

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

Step 3.2.4

Cancel the common factor.

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

Step 3.2.5

Rewrite the expression.

$\int - \frac{1}{2} d x + \int - \frac{2}{- x + 1} d x$

Step 4

Apply the constant rule.

$- \frac{1}{2} x + C + \int - \frac{2}{- x + 1} d x$

Step 5

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

$- \frac{1}{2} x + C - \int \frac{2}{- x + 1} d x$

Step 6

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

$- \frac{1}{2} x + C - (2 \int \frac{1}{- x + 1} d x)$

Step 7

Multiply $2$ by $- 1$ .

$- \frac{1}{2} x + C - 2 \int \frac{1}{- x + 1} d x$

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 $d u$ .

$- \frac{1}{2} x + C - 2 \int \frac{- 1}{u} d u$

Step 9

Move the negative in front of the fraction.

$- \frac{1}{2} x + C - 2 \int - \frac{1}{u} d u$

Step 10

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

$- \frac{1}{2} x + C - 2 (- \int \frac{1}{u} d u)$

Step 11

Multiply $- 1$ by $- 2$ .

$- \frac{1}{2} x + C + 2 \int \frac{1}{u} d u$

Step 12

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

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

Step 13

Simplify.

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

Step 14

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

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