Calculus Examples

\int_{0}^{1} \frac{x - 4}{x^{2} - 5 x + 6} d x

$\int_{0}^{1} \frac{x - 4}{x^{2} - 5 x + 6} d x$

Step 1

Write the fraction using partial fraction decomposition.

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Step 1.1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1.1

Factor

x^{2} - 5 x + 6

$x^{2} - 5 x + 6$ using the AC method.

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Step 1.1.1.1

Consider the form

x^{2} + b x + c

$x^{2} + b x + c$ . Find a pair of integers whose product is

c

$c$ and whose sum is

b

$b$ . In this case, whose product is

6

$6$ and whose sum is

- 5

$- 5$ .

- 3, - 2

$- 3, - 2$

Step 1.1.1.2

Write the factored form using these integers.

\frac{x - 4}{(x - 3) (x - 2)}

$\frac{x - 4}{(x - 3) (x - 2)}$

\frac{x - 4}{(x - 3) (x - 2)}

$\frac{x - 4}{(x - 3) (x - 2)}$

Step 1.1.2

For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor in the denominator is linear, put a single variable in its place

A

$A$ .

\frac{A}{x - 3}

$\frac{A}{x - 3}$

Step 1.1.3

B

$B$ .

\frac{A}{x - 3} + \frac{B}{x - 2}

$\frac{A}{x - 3} + \frac{B}{x - 2}$

Step 1.1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is

(x - 3) (x - 2)

$(x - 3) (x - 2)$ .

\frac{(x - 4) (x - 3) (x - 2)}{(x - 3) (x - 2)} = \frac{(A) (x - 3) (x - 2)}{x - 3} + \frac{(B) (x - 3) (x - 2)}{x - 2}

$\frac{(x - 4) (x - 3) (x - 2)}{(x - 3) (x - 2)} = \frac{(A) (x - 3) (x - 2)}{x - 3} + \frac{(B) (x - 3) (x - 2)}{x - 2}$

Step 1.1.5

Cancel the common factor of

x - 3

$x - 3$ .

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Step 1.1.5.1

Cancel the common factor.

$\frac{(x - 4) (x - 3) (x - 2)}{(x - 3) (x - 2)} = \frac{(A) (x - 3) (x - 2)}{x - 3} + \frac{(B) (x - 3) (x - 2)}{x - 2}$

Step 1.1.5.2

Rewrite the expression.

$\frac{(x - 4) (x - 2)}{x - 2} = \frac{(A) (x - 3) (x - 2)}{x - 3} + \frac{(B) (x - 3) (x - 2)}{x - 2}$

Step 1.1.6

Cancel the common factor of

$x - 2$ .

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Step 1.1.6.1

Cancel the common factor.

$\frac{(x - 4) (x - 2)}{x - 2} = \frac{(A) (x - 3) (x - 2)}{x - 3} + \frac{(B) (x - 3) (x - 2)}{x - 2}$

Step 1.1.6.2

Divide

$x - 4$ by

$1$ .

$x - 4 = \frac{(A) (x - 3) (x - 2)}{x - 3} + \frac{(B) (x - 3) (x - 2)}{x - 2}$

Step 1.1.7

Simplify each term.

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Step 1.1.7.1

Cancel the common factor of

$x - 3$ .

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Step 1.1.7.1.1

Cancel the common factor.

$x - 4 = \frac{A (x - 3) (x - 2)}{x - 3} + \frac{(B) (x - 3) (x - 2)}{x - 2}$

Step 1.1.7.1.2

Divide

$(A) (x - 2)$ by

$1$ .

$x - 4 = (A) (x - 2) + \frac{(B) (x - 3) (x - 2)}{x - 2}$

Step 1.1.7.2

Apply the distributive property.

$x - 4 = A x + A \cdot - 2 + \frac{(B) (x - 3) (x - 2)}{x - 2}$

Step 1.1.7.3

Move

$- 2$ to the left of

$A$ .

$x - 4 = A x - 2 \cdot A + \frac{(B) (x - 3) (x - 2)}{x - 2}$

Step 1.1.7.4

Cancel the common factor of

$x - 2$ .

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Step 1.1.7.4.1

Cancel the common factor.

$x - 4 = A x - 2 A + \frac{(B) (x - 3) (x - 2)}{x - 2}$

Step 1.1.7.4.2

Divide

$(B) (x - 3)$ by

$1$ .

$x - 4 = A x - 2 A + (B) (x - 3)$

Step 1.1.7.5

Apply the distributive property.

$x - 4 = A x - 2 A + B x + B \cdot - 3$

Step 1.1.7.6

Move

$- 3$ to the left of

$B$ .

$x - 4 = A x - 2 A + B x - 3 B$

Step 1.1.8

Move

$- 2 A$ .

$x - 4 = A x + B x - 2 A - 3 B$

Step 1.2

Create equations for the partial fraction variables and use them to set up a system of equations.

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Step 1.2.1

Create an equation for the partial fraction variables by equating the coefficients of

$x$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$1 = A + B$

Step 1.2.2

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing

$x$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$- 4 = - 2 A - 3 B$

Step 1.2.3

Set up the system of equations to find the coefficients of the partial fractions.

$1 = A + B$

$- 4 = - 2 A - 3 B$

$1 = A + B$

$- 4 = - 2 A - 3 B$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Solve for

$A$ in

$1 = A + B$ .

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Step 1.3.1.1

Rewrite the equation as

$A + B = 1$ .

$A + B = 1$

$- 4 = - 2 A - 3 B$

Step 1.3.1.2

Subtract

$B$ from both sides of the equation.

$A = 1 - B$

$- 4 = - 2 A - 3 B$

$A = 1 - B$

$- 4 = - 2 A - 3 B$

Step 1.3.2

Replace all occurrences of

$A$ with

$1 - B$ in each equation.

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Step 1.3.2.1

Replace all occurrences of

$A$ in

$- 4 = - 2 A - 3 B$ with

$1 - B$ .

$- 4 = - 2 (1 - B) - 3 B$

$A = 1 - B$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Simplify

$- 2 (1 - B) - 3 B$ .

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Step 1.3.2.2.1.1

Simplify each term.

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Step 1.3.2.2.1.1.1

Apply the distributive property.

$- 4 = - 2 \cdot 1 - 2 (- B) - 3 B$

$A = 1 - B$

Step 1.3.2.2.1.1.2

Multiply

$- 2$ by

$1$ .

$- 4 = - 2 - 2 (- B) - 3 B$

$A = 1 - B$

Step 1.3.2.2.1.1.3

Multiply

$- 1$ by

$- 2$ .

$- 4 = - 2 + 2 B - 3 B$

$A = 1 - B$

$- 4 = - 2 + 2 B - 3 B$

$A = 1 - B$

Step 1.3.2.2.1.2

Subtract

$3 B$ from

$2 B$ .

$- 4 = - 2 - B$

$A = 1 - B$

$- 4 = - 2 - B$

$A = 1 - B$

$- 4 = - 2 - B$

$A = 1 - B$

$- 4 = - 2 - B$

$A = 1 - B$

Step 1.3.3

Solve for

$B$ in

$- 4 = - 2 - B$ .

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Step 1.3.3.1

Rewrite the equation as

$- 2 - B = - 4$ .

$- 2 - B = - 4$

$A = 1 - B$

Step 1.3.3.2

Move all terms not containing

$B$ to the right side of the equation.

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Step 1.3.3.2.1

Add

$2$ to both sides of the equation.

$- B = - 4 + 2$

$A = 1 - B$

Step 1.3.3.2.2

Add

$- 4$ and

$2$ .

$- B = - 2$

$A = 1 - B$

$- B = - 2$

$A = 1 - B$

Step 1.3.3.3

Divide each term in

$- B = - 2$ by

$- 1$ and simplify.

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Step 1.3.3.3.1

Divide each term in

$- B = - 2$ by

$- 1$ .

$\frac{- B}{- 1} = \frac{- 2}{- 1}$

$A = 1 - B$

Step 1.3.3.3.2

Simplify the left side.

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Step 1.3.3.3.2.1

Dividing two negative values results in a positive value.

$\frac{B}{1} = \frac{- 2}{- 1}$

$A = 1 - B$

Step 1.3.3.3.2.2

Divide

$B$ by

$1$ .

$B = \frac{- 2}{- 1}$

$A = 1 - B$

$B = \frac{- 2}{- 1}$

$A = 1 - B$

Step 1.3.3.3.3

Simplify the right side.

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Step 1.3.3.3.3.1

Divide

$- 2$ by

$- 1$ .

$B = 2$

$A = 1 - B$

$B = 2$

$A = 1 - B$

$B = 2$

$A = 1 - B$

$B = 2$

$A = 1 - B$

Step 1.3.4

Replace all occurrences of

$B$ with

$2$ in each equation.

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Step 1.3.4.1

Replace all occurrences of

$B$ in

$A = 1 - B$ with

$2$ .

$A = 1 - (2)$

$B = 2$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

Simplify

$1 - (2)$ .

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Step 1.3.4.2.1.1

Multiply

$- 1$ by

$2$ .

$A = 1 - 2$

$B = 2$

Step 1.3.4.2.1.2

Subtract

$2$ from

$1$ .

$A = - 1$

$B = 2$

$A = - 1$

$B = 2$

$A = - 1$

$B = 2$

$A = - 1$

$B = 2$

Step 1.3.5

List all of the solutions.

$A = - 1, B = 2$

Step 1.4

Replace each of the partial fraction coefficients in

$\frac{A}{x - 3} + \frac{B}{x - 2}$ with the values found for

$A$ and

$B$ .

$\frac{- 1}{x - 3} + \frac{2}{x - 2}$

Step 1.5

Move the negative in front of the fraction.

$\int_{0}^{1} - \frac{1}{x - 3} + \frac{2}{x - 2} d x$

Step 2

Split the single integral into multiple integrals.

$\int_{0}^{1} - \frac{1}{x - 3} d x + \int_{0}^{1} \frac{2}{x - 2} d x$

Step 3

Since

$- 1$ is constant with respect to

$x$ , move

$- 1$ out of the integral.

$- \int_{0}^{1} \frac{1}{x - 3} d x + \int_{0}^{1} \frac{2}{x - 2} d x$

Step 4

Let

$u_{1} = x - 3$ . Then

$d u_{1} = d x$ . Rewrite using

$u_{1}$ and

$d$

$u_{1}$ .

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Step 4.1

Let

$u_{1} = x - 3$ . Find

$\frac{d u_{1}}{d x}$ .

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Step 4.1.1

Differentiate

$x - 3$ .

$\frac{d}{d x} [x - 3]$

Step 4.1.2

By the Sum Rule, the derivative of

$x - 3$ with respect to

$x$ is

$\frac{d}{d x} [x] + \frac{d}{d x} [- 3]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [- 3]$

Step 4.1.3

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

$1 + \frac{d}{d x} [- 3]$

Step 4.1.4

Since

$- 3$ is constant with respect to

$x$ , the derivative of

$- 3$ with respect to

$x$ is

$0$ .

$1 + 0$

Step 4.1.5

Add

$1$ and

$0$ .

$1$

Step 4.2

Substitute the lower limit in for

$x$ in

$u_{1} = x - 3$ .

$u_{lower} = 0 - 3$

Step 4.3

Subtract

$3$ from

$0$ .

$u_{lower} = - 3$

Step 4.4

Substitute the upper limit in for

$x$ in

$u_{1} = x - 3$ .

$u_{upper} = 1 - 3$

Step 4.5

Subtract

$3$ from

$1$ .

$u_{upper} = - 2$

Step 4.6

The values found for

$u_{lower}$ and

$u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = - 3$

$u_{upper} = - 2$

Step 4.7

Rewrite the problem using

$u_{1}$ ,

$d u_{1}$ , and the new limits of integration.

$- \int_{- 3}^{- 2} \frac{1}{u_{1}} d u_{1} + \int_{0}^{1} \frac{2}{x - 2} d x$

Step 5

The integral of

$\frac{1}{u_{1}}$ with respect to

$u_{1}$ is

$\ln (| u_{1} |)$ .

$- (\ln (| u_{1} |)]_{- 3}^{- 2}) + \int_{0}^{1} \frac{2}{x - 2} d x$

Step 6

Since

$2$ is constant with respect to

$x$ , move

$2$ out of the integral.

$- (\ln (| u_{1} |)]_{- 3}^{- 2}) + 2 \int_{0}^{1} \frac{1}{x - 2} d x$

Step 7

Let

$u_{2} = x - 2$ . Then

$d u_{2} = d x$ . Rewrite using

$u_{2}$ and

$d$

$u_{2}$ .

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Step 7.1

Let

$u_{2} = x - 2$ . Find

$\frac{d u_{2}}{d x}$ .

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Step 7.1.1

Differentiate

$x - 2$ .

$\frac{d}{d x} [x - 2]$

Step 7.1.2

By the Sum Rule, the derivative of

$x - 2$ with respect to

$x$ is

$\frac{d}{d x} [x] + \frac{d}{d x} [- 2]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [- 2]$

Step 7.1.3

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

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

Step 7.1.4

Since

$- 2$ is constant with respect to

$x$ , the derivative of

$- 2$ with respect to

$x$ is

$0$ .

$1 + 0$

Step 7.1.5

Add

$1$ and

$0$ .

$1$

Step 7.2

Substitute the lower limit in for

$x$ in

$u_{2} = x - 2$ .

$u_{lower} = 0 - 2$

Step 7.3

Subtract

$2$ from

$0$ .

$u_{lower} = - 2$

Step 7.4

Substitute the upper limit in for

$x$ in

$u_{2} = x - 2$ .

$u_{upper} = 1 - 2$

Step 7.5

Subtract

$2$ from

$1$ .

$u_{upper} = - 1$

Step 7.6

The values found for

$u_{lower}$ and

$u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = - 2$

$u_{upper} = - 1$

Step 7.7

Rewrite the problem using

$u_{2}$ ,

$d u_{2}$ , and the new limits of integration.

$- (\ln (| u_{1} |)]_{- 3}^{- 2}) + 2 \int_{- 2}^{- 1} \frac{1}{u_{2}} d u_{2}$

Step 8

The integral of

$\frac{1}{u_{2}}$ with respect to

$u_{2}$ is

$\ln (| u_{2} |)$ .

$- (\ln (| u_{1} |)]_{- 3}^{- 2}) + 2 (\ln (| u_{2} |)]_{- 2}^{- 1})$

Step 9

Substitute and simplify.

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Step 9.1

Evaluate

$\ln (| u_{1} |)$ at

$- 2$ and at

$- 3$ .

$- ((\ln (| - 2 |)) - \ln (| - 3 |)) + 2 (\ln (| u_{2} |)]_{- 2}^{- 1})$

Step 9.2

Evaluate

$\ln (| u_{2} |)$ at

$- 1$ and at

$- 2$ .

$- ((\ln (| - 2 |)) - \ln (| - 3 |)) + 2 ((\ln (| - 1 |)) - \ln (| - 2 |))$

Step 9.3

Remove parentheses.

$- (\ln (| - 2 |) - \ln (| - 3 |)) + 2 (\ln (| - 1 |) - \ln (| - 2 |))$

Step 10

Simplify.

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Step 10.1

Use the quotient property of logarithms,

$\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$- \ln (\frac{| - 2 |}{| - 3 |}) + 2 (\ln (| - 1 |) - \ln (| - 2 |))$

Step 10.2

Use the quotient property of logarithms,

$\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$- \ln (\frac{| - 2 |}{| - 3 |}) + 2 \ln (\frac{| - 1 |}{| - 2 |})$

Step 11

Simplify.

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Step 11.1

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

$- 2$ and

$0$ is

$2$ .

$- \ln (\frac{2}{| - 3 |}) + 2 \ln (\frac{| - 1 |}{| - 2 |})$

Step 11.2

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

$- 3$ and

$0$ is

$3$ .

$- \ln (\frac{2}{3}) + 2 \ln (\frac{| - 1 |}{| - 2 |})$

Step 11.3

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

$- 1$ and

$0$ is

$1$ .

$- \ln (\frac{2}{3}) + 2 \ln (\frac{1}{| - 2 |})$

Step 11.4

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

$- 2$ and

$0$ is

$2$ .

$- \ln (\frac{2}{3}) + 2 \ln (\frac{1}{2})$

Step 12

The result can be shown in multiple forms.

Exact Form:

$- \ln (\frac{2}{3}) + 2 \ln (\frac{1}{2})$

Decimal Form:

$- 0.98082925 \dots$

Step 13