Calculus Examples

$\int_{0}^{2} \frac{x}{1 - x} d x$

Step 1

Reorder $1$ and $- x$ .

$\int_{0}^{2} \frac{x}{- x + 1} d x$

Step 2

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

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Step 2.1

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 2.2

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

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

Step 2.3

Multiply the new quotient term by the divisor.

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

Step 2.4

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 2.5

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 2.6

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

$\int_{0}^{2} - 1 + \frac{1}{- x + 1} d x$

Step 3

Split the single integral into multiple integrals.

$\int_{0}^{2} - 1 d x + \int_{0}^{2} \frac{1}{- x + 1} d x$

Step 4

Apply the constant rule.

$- x]_{0}^{2} + \int_{0}^{2} \frac{1}{- x + 1} d x$

Step 5

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 5.1

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

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Step 5.1.1

Rewrite.

$\frac{- 1}{1}$

Step 5.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 5.2

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

$u_{lower} = - 0 + 1$

Step 5.3

Simplify.

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Step 5.3.1

Multiply $- 1$ by $0$ .

$u_{lower} = 0 + 1$

Step 5.3.2

Add $0$ and $1$ .

$u_{lower} = 1$

Step 5.4

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

$u_{upper} = - 1 \cdot 2 + 1$

Step 5.5

Simplify.

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Step 5.5.1

Multiply $- 1$ by $2$ .

$u_{upper} = - 2 + 1$

Step 5.5.2

Add $- 2$ and $1$ .

$u_{upper} = - 1$

Step 5.6

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

$u_{lower} = 1$

$u_{upper} = - 1$

Step 5.7

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

$- x]_{0}^{2} + \int_{1}^{- 1} \frac{- 1}{u} d u$

Step 6

Move the negative in front of the fraction.

$- x]_{0}^{2} + \int_{1}^{- 1} - \frac{1}{u} d u$

Step 7

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

$- x]_{0}^{2} - \int_{1}^{- 1} \frac{1}{u} d u$

Step 8

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

$- x]_{0}^{2} - (\ln (| u |)]_{1}^{- 1})$

Step 9

Substitute and simplify.

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Step 9.1

Evaluate $- x$ at $2$ and at $0$ .

$(- 1 \cdot 2) + 0 - (\ln (| u |)]_{1}^{- 1})$

Step 9.2

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

$(- 1 \cdot 2) + 0 - ((\ln (| - 1 |)) - \ln (| 1 |))$

Step 9.3

Simplify.

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Step 9.3.1

Multiply $- 1$ by $2$ .

$- 2 + 0 - ((\ln (| - 1 |)) - \ln (| 1 |))$

Step 9.3.2

Add $- 2$ and $0$ .

$- 2 - (\ln (| - 1 |) - \ln (| 1 |))$

Step 10

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

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

Step 11

Simplify.

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Step 11.1

The absolute value is the distance between a number and zero. The distance between $- 1$ and $0$ is $1$ .

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

Step 11.2

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$- 2 - \ln (\frac{1}{1})$

Step 11.3

Divide $1$ by $1$ .

$- 2 - \ln (1)$

Step 11.4

The natural logarithm of $1$ is $0$ .

$- 2 - 0$

Step 11.5

Multiply $- 1$ by $0$ .

$- 2 + 0$

Step 11.6

Add $- 2$ and $0$ .

$- 2$

Step 12