Calculus Examples

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

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

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} + x - 2

$x^{2} + x - 2$ 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

- 2

$- 2$ and whose sum is

1

$1$ .

- 1, 2

$- 1, 2$

Step 1.1.1.2

Write the factored form using these integers.

\frac{x + 5}{(x - 1) (x + 2)}

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

\frac{x + 5}{(x - 1) (x + 2)}

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

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 - 1}

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

Step 1.1.3

B

$B$ .

\frac{A}{x - 1} + \frac{B}{x + 2}

$\frac{A}{x - 1} + \frac{B}{x + 2}$

Step 1.1.4

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

(x - 1) (x + 2)

$(x - 1) (x + 2)$ .

\frac{(x + 5) (x - 1) (x + 2)}{(x - 1) (x + 2)} = \frac{(A) (x - 1) (x + 2)}{x - 1} + \frac{(B) (x - 1) (x + 2)}{x + 2}

$\frac{(x + 5) (x - 1) (x + 2)}{(x - 1) (x + 2)} = \frac{(A) (x - 1) (x + 2)}{x - 1} + \frac{(B) (x - 1) (x + 2)}{x + 2}$

Step 1.1.5

Cancel the common factor of

x - 1

$x - 1$ .

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Step 1.1.5.1

Cancel the common factor.

$\frac{(x + 5) (x - 1) (x + 2)}{(x - 1) (x + 2)} = \frac{(A) (x - 1) (x + 2)}{x - 1} + \frac{(B) (x - 1) (x + 2)}{x + 2}$

Step 1.1.5.2

Rewrite the expression.

$\frac{(x + 5) (x + 2)}{x + 2} = \frac{(A) (x - 1) (x + 2)}{x - 1} + \frac{(B) (x - 1) (x + 2)}{x + 2}$

Step 1.1.6

Cancel the common factor of

$x + 2$ .

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Step 1.1.6.1

Cancel the common factor.

$\frac{(x + 5) (x + 2)}{x + 2} = \frac{(A) (x - 1) (x + 2)}{x - 1} + \frac{(B) (x - 1) (x + 2)}{x + 2}$

Step 1.1.6.2

Divide

$x + 5$ by

$1$ .

$x + 5 = \frac{(A) (x - 1) (x + 2)}{x - 1} + \frac{(B) (x - 1) (x + 2)}{x + 2}$

Step 1.1.7

Simplify each term.

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Step 1.1.7.1

Cancel the common factor of

$x - 1$ .

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Step 1.1.7.1.1

Cancel the common factor.

$x + 5 = \frac{A (x - 1) (x + 2)}{x - 1} + \frac{(B) (x - 1) (x + 2)}{x + 2}$

Step 1.1.7.1.2

Divide

$(A) (x + 2)$ by

$1$ .

$x + 5 = (A) (x + 2) + \frac{(B) (x - 1) (x + 2)}{x + 2}$

Step 1.1.7.2

Apply the distributive property.

$x + 5 = A x + A \cdot 2 + \frac{(B) (x - 1) (x + 2)}{x + 2}$

Step 1.1.7.3

Move

$2$ to the left of

$A$ .

$x + 5 = A x + 2 \cdot A + \frac{(B) (x - 1) (x + 2)}{x + 2}$

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.

$x + 5 = A x + 2 A + \frac{(B) (x - 1) (x + 2)}{x + 2}$

Step 1.1.7.4.2

Divide

$(B) (x - 1)$ by

$1$ .

$x + 5 = A x + 2 A + (B) (x - 1)$

Step 1.1.7.5

Apply the distributive property.

$x + 5 = A x + 2 A + B x + B \cdot - 1$

Step 1.1.7.6

Move

$- 1$ to the left of

$B$ .

$x + 5 = A x + 2 A + B x - 1 \cdot B$

Step 1.1.7.7

Rewrite

$- 1 B$ as

$- B$ .

$x + 5 = A x + 2 A + B x - B$

Step 1.1.8

Move

$2 A$ .

$x + 5 = A x + B x + 2 A - B$

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.

$1 = A + B$

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.

$5 = 2 A - 1 B$

Step 1.2.3

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

$1 = A + B$

$5 = 2 A - 1 B$

$1 = A + B$

$5 = 2 A - 1 B$

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

$A + B = 1$ .

$A + B = 1$

$5 = 2 A - 1 B$

Step 1.3.1.2

Subtract

$B$ from both sides of the equation.

$A = 1 - B$

$5 = 2 A - 1 B$

$A = 1 - B$

$5 = 2 A - 1 B$

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

$5 = 2 A - 1 B$ with

$1 - B$ .

$5 = 2 (1 - B) - 1 B$

$A = 1 - B$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Simplify

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

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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.

$5 = 2 \cdot 1 + 2 (- B) - 1 B$

$A = 1 - B$

Step 1.3.2.2.1.1.2

Multiply

$2$ by

$1$ .

$5 = 2 + 2 (- B) - 1 B$

$A = 1 - B$

Step 1.3.2.2.1.1.3

Multiply

$- 1$ by

$2$ .

$5 = 2 - 2 B - 1 B$

$A = 1 - B$

Step 1.3.2.2.1.1.4

Rewrite

$- 1 B$ as

$- B$ .

$5 = 2 - 2 B - B$

$A = 1 - B$

$5 = 2 - 2 B - B$

$A = 1 - B$

Step 1.3.2.2.1.2

Subtract

$B$ from

$- 2 B$ .

$5 = 2 - 3 B$

$A = 1 - B$

$5 = 2 - 3 B$

$A = 1 - B$

$5 = 2 - 3 B$

$A = 1 - B$

$5 = 2 - 3 B$

$A = 1 - B$

Step 1.3.3

Solve for

$B$ in

$5 = 2 - 3 B$ .

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Step 1.3.3.1

Rewrite the equation as

$2 - 3 B = 5$ .

$2 - 3 B = 5$

$A = 1 - B$

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

Subtract

$2$ from both sides of the equation.

$- 3 B = 5 - 2$

$A = 1 - B$

Step 1.3.3.2.2

Subtract

$2$ from

$5$ .

$- 3 B = 3$

$A = 1 - B$

$- 3 B = 3$

$A = 1 - B$

Step 1.3.3.3

Divide each term in

$- 3 B = 3$ by

$- 3$ and simplify.

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Step 1.3.3.3.1

Divide each term in

$- 3 B = 3$ by

$- 3$ .

$\frac{- 3 B}{- 3} = \frac{3}{- 3}$

$A = 1 - B$

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

$- 3$ .

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Step 1.3.3.3.2.1.1

Cancel the common factor.

$\frac{- 3 B}{- 3} = \frac{3}{- 3}$

$A = 1 - B$

Step 1.3.3.3.2.1.2

Divide

$B$ by

$1$ .

$B = \frac{3}{- 3}$

$A = 1 - B$

$B = \frac{3}{- 3}$

$A = 1 - B$

$B = \frac{3}{- 3}$

$A = 1 - B$

Step 1.3.3.3.3

Simplify the right side.

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Step 1.3.3.3.3.1

Divide

$3$ by

$- 3$ .

$B = - 1$

$A = 1 - B$

$B = - 1$

$A = 1 - B$

$B = - 1$

$A = 1 - B$

$B = - 1$

$A = 1 - B$

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

$A = 1 - B$ with

$- 1$ .

$A = 1 - (- 1)$

$B = - 1$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

Simplify

$1 - (- 1)$ .

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Step 1.3.4.2.1.1

Multiply

$- 1$ by

$- 1$ .

$A = 1 + 1$

$B = - 1$

Step 1.3.4.2.1.2

Add

$1$ and

$1$ .

$A = 2$

$B = - 1$

$A = 2$

$B = - 1$

$A = 2$

$B = - 1$

$A = 2$

$B = - 1$

Step 1.3.5

List all of the solutions.

$A = 2, B = - 1$

Step 1.4

Replace each of the partial fraction coefficients in

$\frac{A}{x - 1} + \frac{B}{x + 2}$ with the values found for

$A$ and

$B$ .

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

Step 1.5

Move the negative in front of the fraction.

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

Step 2

Split the single integral into multiple integrals.

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

Step 3

Since

$2$ is constant with respect to

$x$ , move

$2$ out of the integral.

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

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

Rewrite the problem using

$u_{1}$ and

$d u_{1}$ .

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

Step 5

The integral of

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

$u_{1}$ is

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

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

Step 6

Since

$- 1$ is constant with respect to

$x$ , move

$- 1$ out of the integral.

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

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

Rewrite the problem using

$u_{2}$ and

$d u_{2}$ .

$2 (\ln (| u_{1} |) + C) - \int \frac{1}{u_{2}} d u_{2}$

Step 8

The integral of

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

$u_{2}$ is

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

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

Step 9

Simplify.

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

Step 10

Substitute back in for each integration substitution variable.

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Step 10.1

Replace all occurrences of

$u_{1}$ with

$x - 1$ .

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

Step 10.2

Replace all occurrences of

$u_{2}$ with

$x + 2$ .

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

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