Calculus Examples

$\int_{3}^{5} \frac{3 x + 5}{x^{2} + 2 x - 3} d x$

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 $x^{2} + 2 x - 3$ using the AC method.

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Step 1.1.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $2$ .

$- 1, 3$

Step 1.1.1.2

Write the factored form using these integers.

$\frac{3 x + 5}{(x - 1) (x + 3)}$

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 $A$ .

$\frac{A}{x - 1}$

Step 1.1.3

$\frac{A}{x - 1} + \frac{B}{x + 3}$

Step 1.1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(x - 1) (x + 3)$ .

$\frac{(3 x + 5) (x - 1) (x + 3)}{(x - 1) (x + 3)} = \frac{(A) (x - 1) (x + 3)}{x - 1} + \frac{(B) (x - 1) (x + 3)}{x + 3}$

Step 1.1.5

Cancel the common factor of $x - 1$ .

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Step 1.1.5.1

Cancel the common factor.

$\frac{(3 x + 5) (x - 1) (x + 3)}{(x - 1) (x + 3)} = \frac{(A) (x - 1) (x + 3)}{x - 1} + \frac{(B) (x - 1) (x + 3)}{x + 3}$

Step 1.1.5.2

Rewrite the expression.

$\frac{(3 x + 5) (x + 3)}{x + 3} = \frac{(A) (x - 1) (x + 3)}{x - 1} + \frac{(B) (x - 1) (x + 3)}{x + 3}$

Step 1.1.6

Cancel the common factor of $x + 3$ .

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Step 1.1.6.1

Cancel the common factor.

$\frac{(3 x + 5) (x + 3)}{x + 3} = \frac{(A) (x - 1) (x + 3)}{x - 1} + \frac{(B) (x - 1) (x + 3)}{x + 3}$

Step 1.1.6.2

Divide $3 x + 5$ by $1$ .

$3 x + 5 = \frac{(A) (x - 1) (x + 3)}{x - 1} + \frac{(B) (x - 1) (x + 3)}{x + 3}$

Step 1.1.7

Simplify each term.

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Step 1.1.7.1

Cancel the common factor of $x - 1$ .

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Step 1.1.7.1.1

Cancel the common factor.

$3 x + 5 = \frac{A (x - 1) (x + 3)}{x - 1} + \frac{(B) (x - 1) (x + 3)}{x + 3}$

Step 1.1.7.1.2

Divide $(A) (x + 3)$ by $1$ .

$3 x + 5 = (A) (x + 3) + \frac{(B) (x - 1) (x + 3)}{x + 3}$

Step 1.1.7.2

Apply the distributive property.

$3 x + 5 = A x + A \cdot 3 + \frac{(B) (x - 1) (x + 3)}{x + 3}$

Step 1.1.7.3

Move $3$ to the left of $A$ .

$3 x + 5 = A x + 3 \cdot A + \frac{(B) (x - 1) (x + 3)}{x + 3}$

Step 1.1.7.4

Cancel the common factor of $x + 3$ .

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Step 1.1.7.4.1

Cancel the common factor.

$3 x + 5 = A x + 3 A + \frac{(B) (x - 1) (x + 3)}{x + 3}$

Step 1.1.7.4.2

Divide $(B) (x - 1)$ by $1$ .

$3 x + 5 = A x + 3 A + (B) (x - 1)$

Step 1.1.7.5

Apply the distributive property.

$3 x + 5 = A x + 3 A + B x + B \cdot - 1$

Step 1.1.7.6

Move $- 1$ to the left of $B$ .

$3 x + 5 = A x + 3 A + B x - 1 \cdot B$

Step 1.1.7.7

Rewrite $- 1 B$ as $- B$ .

$3 x + 5 = A x + 3 A + B x - B$

Step 1.1.8

Move $3 A$ .

$3 x + 5 = A x + B x + 3 A - B$

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.

$3 = A + B$

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.

$5 = 3 A - 1 B$

Step 1.2.3

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

$3 = A + B$

$5 = 3 A - 1 B$

$3 = A + B$

$5 = 3 A - 1 B$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Solve for $A$ in $3 = A + B$ .

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Step 1.3.1.1

Rewrite the equation as $A + B = 3$ .

$A + B = 3$

$5 = 3 A - 1 B$

Step 1.3.1.2

Subtract $B$ from both sides of the equation.

$A = 3 - B$

$5 = 3 A - 1 B$

$A = 3 - B$

$5 = 3 A - 1 B$

Step 1.3.2

Replace all occurrences of $A$ with $3 - B$ in each equation.

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Step 1.3.2.1

Replace all occurrences of $A$ in $5 = 3 A - 1 B$ with $3 - B$ .

$5 = 3 (3 - B) - 1 B$

$A = 3 - B$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

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

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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.

$5 = 3 \cdot 3 + 3 (- B) - 1 B$

$A = 3 - B$

Step 1.3.2.2.1.1.2

Multiply $3$ by $3$ .

$5 = 9 + 3 (- B) - 1 B$

$A = 3 - B$

Step 1.3.2.2.1.1.3

Multiply $- 1$ by $3$ .

$5 = 9 - 3 B - 1 B$

$A = 3 - B$

Step 1.3.2.2.1.1.4

Rewrite $- 1 B$ as $- B$ .

$5 = 9 - 3 B - B$

$A = 3 - B$

$5 = 9 - 3 B - B$

$A = 3 - B$

Step 1.3.2.2.1.2

Subtract $B$ from $- 3 B$ .

$5 = 9 - 4 B$

$A = 3 - B$

$5 = 9 - 4 B$

$A = 3 - B$

$5 = 9 - 4 B$

$A = 3 - B$

$5 = 9 - 4 B$

$A = 3 - B$

Step 1.3.3

Solve for $B$ in $5 = 9 - 4 B$ .

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Step 1.3.3.1

Rewrite the equation as $9 - 4 B = 5$ .

$9 - 4 B = 5$

$A = 3 - B$

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

Subtract $9$ from both sides of the equation.

$- 4 B = 5 - 9$

$A = 3 - B$

Step 1.3.3.2.2

Subtract $9$ from $5$ .

$- 4 B = - 4$

$A = 3 - B$

$- 4 B = - 4$

$A = 3 - B$

Step 1.3.3.3

Divide each term in $- 4 B = - 4$ by $- 4$ and simplify.

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Step 1.3.3.3.1

Divide each term in $- 4 B = - 4$ by $- 4$ .

$\frac{- 4 B}{- 4} = \frac{- 4}{- 4}$

$A = 3 - B$

Step 1.3.3.3.2

Simplify the left side.

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Step 1.3.3.3.2.1

Cancel the common factor of $- 4$ .

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Step 1.3.3.3.2.1.1

Cancel the common factor.

$\frac{- 4 B}{- 4} = \frac{- 4}{- 4}$

$A = 3 - B$

Step 1.3.3.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{- 4}{- 4}$

$A = 3 - B$

$B = \frac{- 4}{- 4}$

$A = 3 - B$

$B = \frac{- 4}{- 4}$

$A = 3 - B$

Step 1.3.3.3.3

Simplify the right side.

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Step 1.3.3.3.3.1

Divide $- 4$ by $- 4$ .

$B = 1$

$A = 3 - B$

$B = 1$

$A = 3 - B$

$B = 1$

$A = 3 - B$

$B = 1$

$A = 3 - B$

Step 1.3.4

Replace all occurrences of $B$ with $1$ in each equation.

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Step 1.3.4.1

Replace all occurrences of $B$ in $A = 3 - B$ with $1$ .

$A = 3 - (1)$

$B = 1$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

Simplify $3 - (1)$ .

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Step 1.3.4.2.1.1

Multiply $- 1$ by $1$ .

$A = 3 - 1$

$B = 1$

Step 1.3.4.2.1.2

Subtract $1$ from $3$ .

$A = 2$

$B = 1$

$A = 2$

$B = 1$

$A = 2$

$B = 1$

$A = 2$

$B = 1$

Step 1.3.5

List all of the solutions.

$A = 2, B = 1$

Step 1.4

Replace each of the partial fraction coefficients in $\frac{A}{x - 1} + \frac{B}{x + 3}$ with the values found for $A$ and $B$ .

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

Step 1.5

Remove the zero from the expression.

$\int_{3}^{5} \frac{2}{x - 1} + \frac{1}{x + 3} d x$

Step 2

Split the single integral into multiple integrals.

$\int_{3}^{5} \frac{2}{x - 1} d x + \int_{3}^{5} \frac{1}{x + 3} d x$

Step 3

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

$2 \int_{3}^{5} \frac{1}{x - 1} d x + \int_{3}^{5} \frac{1}{x + 3} d x$

Step 4

Let $u_{1} = x - 1$ . Then $d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 4.1

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

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Step 4.1.1

Differentiate $x - 1$ .

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

Step 4.1.2

By the Sum Rule, the derivative of $x - 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 1]$ .

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

Step 4.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

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

Step 4.1.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ 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 $u_{1} = x - 1$ .

$u_{lower} = 3 - 1$

Step 4.3

Subtract $1$ from $3$ .

$u_{lower} = 2$

Step 4.4

Substitute the upper limit in for $x$ in $u_{1} = x - 1$ .

$u_{upper} = 5 - 1$

Step 4.5

Subtract $1$ from $5$ .

$u_{upper} = 4$

Step 4.6

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

$u_{lower} = 2$

$u_{upper} = 4$

Step 4.7

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

$2 \int_{2}^{4} \frac{1}{u_{1}} d u_{1} + \int_{3}^{5} \frac{1}{x + 3} d x$

Step 5

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

$2 (\ln (| u_{1} |)]_{2}^{4}) + \int_{3}^{5} \frac{1}{x + 3} d x$

Step 6

Let $u_{2} = x + 3$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 6.1

Let $u_{2} = x + 3$ . Find $\frac{d u_{2}}{d x}$ .

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Step 6.1.1

Differentiate $x + 3$ .

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

Step 6.1.2

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 6.1.3

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 6.1.4

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$1 + 0$

Step 6.1.5

Add $1$ and $0$ .

$1$

Step 6.2

Substitute the lower limit in for $x$ in $u_{2} = x + 3$ .

$u_{lower} = 3 + 3$

Step 6.3

Add $3$ and $3$ .

$u_{lower} = 6$

Step 6.4

Substitute the upper limit in for $x$ in $u_{2} = x + 3$ .

$u_{upper} = 5 + 3$

Step 6.5

Add $5$ and $3$ .

$u_{upper} = 8$

Step 6.6

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

$u_{lower} = 6$

$u_{upper} = 8$

Step 6.7

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

$2 (\ln (| u_{1} |)]_{2}^{4}) + \int_{6}^{8} \frac{1}{u_{2}} d u_{2}$

Step 7

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$2 (\ln (| u_{1} |)]_{2}^{4}) + \ln (| u_{2} |)]_{6}^{8}$

Step 8

Substitute and simplify.

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Step 8.1

Evaluate $\ln (| u_{1} |)$ at $4$ and at $2$ .

$2 ((\ln (| 4 |)) - \ln (| 2 |)) + \ln (| u_{2} |)]_{6}^{8}$

Step 8.2

Evaluate $\ln (| u_{2} |)$ at $8$ and at $6$ .

$2 ((\ln (| 4 |)) - \ln (| 2 |)) + (\ln (| 8 |)) - \ln (| 6 |)$

Step 8.3

Remove unnecessary parentheses.

$2 (\ln (| 4 |) - \ln (| 2 |)) + \ln (| 8 |) - \ln (| 6 |)$

Step 9

Simplify.

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Step 9.1

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

$2 \ln (\frac{| 4 |}{| 2 |}) + \ln (| 8 |) - \ln (| 6 |)$

Step 9.2

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

$2 \ln (\frac{| 4 |}{| 2 |}) + \ln (\frac{| 8 |}{| 6 |})$

Step 10

Simplify.

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Step 10.1

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$2 \ln (\frac{4}{| 2 |}) + \ln (\frac{| 8 |}{| 6 |})$

Step 10.2

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$2 \ln (\frac{4}{2}) + \ln (\frac{| 8 |}{| 6 |})$

Step 10.3

Divide $4$ by $2$ .

$2 \ln (2) + \ln (\frac{| 8 |}{| 6 |})$

Step 10.4

The absolute value is the distance between a number and zero. The distance between $0$ and $8$ is $8$ .

$2 \ln (2) + \ln (\frac{8}{| 6 |})$

Step 10.5

The absolute value is the distance between a number and zero. The distance between $0$ and $6$ is $6$ .

$2 \ln (2) + \ln (\frac{8}{6})$

Step 10.6

Cancel the common factor of $8$ and $6$ .

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Step 10.6.1

Factor $2$ out of $8$ .

$2 \ln (2) + \ln (\frac{2 (4)}{6})$

Step 10.6.2

Cancel the common factors.

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Step 10.6.2.1

Factor $2$ out of $6$ .

$2 \ln (2) + \ln (\frac{2 \cdot 4}{2 \cdot 3})$

Step 10.6.2.2

Cancel the common factor.

$2 \ln (2) + \ln (\frac{2 \cdot 4}{2 \cdot 3})$

Step 10.6.2.3

Rewrite the expression.

$2 \ln (2) + \ln (\frac{4}{3})$

Step 11

The result can be shown in multiple forms.

Exact Form:

$2 \ln (2) + \ln (\frac{4}{3})$

Decimal Form:

$1.67397643 \dots$

Step 12