Calculus Examples

$\int \frac{7 x^{5} + 7 x^{2} + 14}{x^{3} - x} d x$

Step 1

Divide $7 x^{5} + 7 x^{2} + 14$ by $x^{3} - x$ .

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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 $0$ .

$x^{3}$

$0 x^{2}$

$x$

$0$

$7 x^{5}$

$0 x^{4}$

$0 x^{3}$

$7 x^{2}$

$0 x$

$14$

Step 1.2

Divide the highest order term in the dividend $7 x^{5}$ by the highest order term in divisor $x^{3}$ .

$7 x^{2}$

$x^{3}$

$0 x^{2}$

$x$

$0$

$7 x^{5}$

$0 x^{4}$

$0 x^{3}$

$7 x^{2}$

$0 x$

$14$

Step 1.3

Multiply the new quotient term by the divisor.

$7 x^{2}$

$x^{3}$

$0 x^{2}$

$x$

$0$

$7 x^{5}$

$0 x^{4}$

$0 x^{3}$

$7 x^{2}$

$0 x$

$14$

$7 x^{5}$

$0$

$7 x^{3}$

$0$

Step 1.4

The expression needs to be subtracted from the dividend, so change all the signs in $7 x^{5} + 0 - 7 x^{3} + 0$

$7 x^{2}$

$x^{3}$

$0 x^{2}$

$x$

$0$

$7 x^{5}$

$0 x^{4}$

$0 x^{3}$

$7 x^{2}$

$0 x$

$14$

$7 x^{5}$

$0$

$7 x^{3}$

$0$

Step 1.5

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

$7 x^{2}$

$x^{3}$

$0 x^{2}$

$x$

$0$

$7 x^{5}$

$0 x^{4}$

$0 x^{3}$

$7 x^{2}$

$0 x$

$14$

$7 x^{5}$

$0$

$7 x^{3}$

$0$

$7 x^{3}$

$7 x^{2}$

Step 1.6

Pull the next term from the original dividend down into the current dividend.

$7 x^{2}$

$x^{3}$

$0 x^{2}$

$x$

$0$

$7 x^{5}$

$0 x^{4}$

$0 x^{3}$

$7 x^{2}$

$0 x$

$14$

$7 x^{5}$

$0$

$7 x^{3}$

$0$

$7 x^{3}$

$7 x^{2}$

$0 x$

$14$

Step 1.7

Divide the highest order term in the dividend $7 x^{3}$ by the highest order term in divisor $x^{3}$ .

$7 x^{2}$

$0 x$

$7$

$x^{3}$

$0 x^{2}$

$x$

$0$

$7 x^{5}$

$0 x^{4}$

$0 x^{3}$

$7 x^{2}$

$0 x$

$14$

$7 x^{5}$

$0$

$7 x^{3}$

$0$

$7 x^{3}$

$7 x^{2}$

$0 x$

$14$

Step 1.8

Multiply the new quotient term by the divisor.

$7 x^{2}$

$0 x$

$7$

$x^{3}$

$0 x^{2}$

$x$

$0$

$7 x^{5}$

$0 x^{4}$

$0 x^{3}$

$7 x^{2}$

$0 x$

$14$

$7 x^{5}$

$0$

$7 x^{3}$

$0$

$7 x^{3}$

$7 x^{2}$

$0 x$

$14$

$7 x^{3}$

$0$

$7 x$

$0$

Step 1.9

The expression needs to be subtracted from the dividend, so change all the signs in $7 x^{3} + 0 - 7 x + 0$

$7 x^{2}$

$0 x$

$7$

$x^{3}$

$0 x^{2}$

$x$

$0$

$7 x^{5}$

$0 x^{4}$

$0 x^{3}$

$7 x^{2}$

$0 x$

$14$

$7 x^{5}$

$0$

$7 x^{3}$

$0$

$7 x^{3}$

$7 x^{2}$

$0 x$

$14$

$7 x^{3}$

$0$

$7 x$

$0$

Step 1.10

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

$7 x^{2}$

$0 x$

$7$

$x^{3}$

$0 x^{2}$

$x$

$0$

$7 x^{5}$

$0 x^{4}$

$0 x^{3}$

$7 x^{2}$

$0 x$

$14$

$7 x^{5}$

$0$

$7 x^{3}$

$0$

$7 x^{3}$

$7 x^{2}$

$0 x$

$14$

$7 x^{3}$

$0$

$7 x$

$0$

$7 x^{2}$

$7 x$

$14$

Step 1.11

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

$\int 7 x^{2} + 7 + \frac{7 x^{2} + 7 x + 14}{x^{3} - x} d x$

Step 2

Split the single integral into multiple integrals.

$\int 7 x^{2} d x + \int 7 d x + \int \frac{7 x^{2} + 7 x + 14}{x^{3} - x} d x$

Step 3

Since $7$ is constant with respect to $x$ , move $7$ out of the integral.

$7 \int x^{2} d x + \int 7 d x + \int \frac{7 x^{2} + 7 x + 14}{x^{3} - x} d x$

Step 4

By the Power Rule, the integral of $x^{2}$ with respect to $x$ is $\frac{1}{3} x^{3}$ .

$7 (\frac{1}{3} x^{3} + C) + \int 7 d x + \int \frac{7 x^{2} + 7 x + 14}{x^{3} - x} d x$

Step 5

Apply the constant rule.

$7 (\frac{1}{3} x^{3} + C) + 7 x + C + \int \frac{7 x^{2} + 7 x + 14}{x^{3} - x} d x$

Step 6

Combine $\frac{1}{3}$ and $x^{3}$ .

$7 (\frac{x^{3}}{3} + C) + 7 x + C + \int \frac{7 x^{2} + 7 x + 14}{x^{3} - x} d x$

Step 7

Write the fraction using partial fraction decomposition.

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Step 7.1

Decompose the fraction and multiply through by the common denominator.

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Step 7.1.1

Factor the fraction.

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Step 7.1.1.1

Factor $7$ out of $7 x^{2} + 7 x + 14$ .

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Step 7.1.1.1.1

Factor $7$ out of $7 x^{2}$ .

$\frac{7 (x^{2}) + 7 x + 14}{x^{3} - x}$

Step 7.1.1.1.2

Factor $7$ out of $7 x$ .

$\frac{7 (x^{2}) + 7 (x) + 14}{x^{3} - x}$

Step 7.1.1.1.3

Factor $7$ out of $14$ .

$\frac{7 (x^{2}) + 7 x + 7 \cdot 2}{x^{3} - x}$

Step 7.1.1.1.4

Factor $7$ out of $7 (x^{2}) + 7 x$ .

$\frac{7 (x^{2} + x) + 7 \cdot 2}{x^{3} - x}$

Step 7.1.1.1.5

Factor $7$ out of $7 (x^{2} + x) + 7 \cdot 2$ .

$\frac{7 (x^{2} + x + 2)}{x^{3} - x}$

Step 7.1.1.2

Factor $x$ out of $x^{3} - x$ .

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Step 7.1.1.2.1

Factor $x$ out of $x^{3}$ .

$\frac{7 (x^{2} + x + 2)}{x \cdot x^{2} - x}$

Step 7.1.1.2.2

Factor $x$ out of $- x$ .

$\frac{7 (x^{2} + x + 2)}{x \cdot x^{2} + x \cdot - 1}$

Step 7.1.1.2.3

Factor $x$ out of $x \cdot x^{2} + x \cdot - 1$ .

$\frac{7 (x^{2} + x + 2)}{x (x^{2} - 1)}$

Step 7.1.1.3

Rewrite $1$ as $1^{2}$ .

$\frac{7 (x^{2} + x + 2)}{x (x^{2} - 1^{2})}$

Step 7.1.1.4

Factor.

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Step 7.1.1.4.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$\frac{7 (x^{2} + x + 2)}{x ((x + 1) (x - 1))}$

Step 7.1.1.4.2

Remove unnecessary parentheses.

$\frac{7 (x^{2} + x + 2)}{x (x + 1) (x - 1)}$

Step 7.1.2

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 $B$ .

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

Step 7.1.3

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

Step 7.1.4

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

$\frac{7 (x^{2} + x + 2) (x (x + 1) (x - 1))}{x (x + 1) (x - 1)} = \frac{A (x (x + 1) (x - 1))}{x} + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.5

Cancel the common factor of $x$ .

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Step 7.1.5.1

Cancel the common factor.

$\frac{7 (x^{2} + x + 2) (x (x + 1) (x - 1))}{x (x + 1) (x - 1)} = \frac{A (x (x + 1) (x - 1))}{x} + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.5.2

Rewrite the expression.

$\frac{7 (x^{2} + x + 2) ((x + 1) (x - 1))}{(x + 1) (x - 1)} = \frac{A (x (x + 1) (x - 1))}{x} + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.6

Cancel the common factor of $x + 1$ .

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Step 7.1.6.1

Cancel the common factor.

$\frac{7 (x^{2} + x + 2) ((x + 1) (x - 1))}{(x + 1) (x - 1)} = \frac{A (x (x + 1) (x - 1))}{x} + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.6.2

Rewrite the expression.

$\frac{7 (x^{2} + x + 2) (x - 1)}{x - 1} = \frac{A (x (x + 1) (x - 1))}{x} + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.7

Cancel the common factor of $x - 1$ .

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Step 7.1.7.1

Cancel the common factor.

$\frac{7 (x^{2} + x + 2) (x - 1)}{x - 1} = \frac{A (x (x + 1) (x - 1))}{x} + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.7.2

Divide $7 (x^{2} + x + 2)$ by $1$ .

$7 (x^{2} + x + 2) = \frac{A (x (x + 1) (x - 1))}{x} + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.8

Apply the distributive property.

$7 x^{2} + 7 x + 7 \cdot 2 = \frac{A (x (x + 1) (x - 1))}{x} + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.9

Multiply $7$ by $2$ .

$7 x^{2} + 7 x + 14 = \frac{A (x (x + 1) (x - 1))}{x} + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10

Simplify each term.

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Step 7.1.10.1

Cancel the common factor of $x$ .

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Step 7.1.10.1.1

Cancel the common factor.

$7 x^{2} + 7 x + 14 = \frac{A (x (x + 1) (x - 1))}{x} + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.1.2

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

$7 x^{2} + 7 x + 14 = A ((x + 1) (x - 1)) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.2

Expand $(x + 1) (x - 1)$ using the FOIL Method.

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Step 7.1.10.2.1

Apply the distributive property.

$7 x^{2} + 7 x + 14 = A (x (x - 1) + 1 (x - 1)) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.2.2

Apply the distributive property.

$7 x^{2} + 7 x + 14 = A (x \cdot x + x \cdot - 1 + 1 (x - 1)) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.2.3

Apply the distributive property.

$7 x^{2} + 7 x + 14 = A (x \cdot x + x \cdot - 1 + 1 x + 1 \cdot - 1) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.3

Simplify and combine like terms.

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Step 7.1.10.3.1

Simplify each term.

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Step 7.1.10.3.1.1

Multiply $x$ by $x$ .

$7 x^{2} + 7 x + 14 = A (x^{2} + x \cdot - 1 + 1 x + 1 \cdot - 1) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.3.1.2

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

$7 x^{2} + 7 x + 14 = A (x^{2} - 1 \cdot x + 1 x + 1 \cdot - 1) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.3.1.3

Rewrite $- 1 x$ as $- x$ .

$7 x^{2} + 7 x + 14 = A (x^{2} - x + 1 x + 1 \cdot - 1) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.3.1.4

Multiply $x$ by $1$ .

$7 x^{2} + 7 x + 14 = A (x^{2} - x + x + 1 \cdot - 1) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.3.1.5

Multiply $- 1$ by $1$ .

$7 x^{2} + 7 x + 14 = A (x^{2} - x + x - 1) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.3.2

Add $- x$ and $x$ .

$7 x^{2} + 7 x + 14 = A (x^{2} + 0 - 1) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.3.3

Add $x^{2}$ and $0$ .

$7 x^{2} + 7 x + 14 = A (x^{2} - 1) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.4

Apply the distributive property.

$7 x^{2} + 7 x + 14 = A x^{2} + A \cdot - 1 + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.5

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

$7 x^{2} + 7 x + 14 = A x^{2} - 1 \cdot A + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.6

Rewrite $- 1 A$ as $- A$ .

$7 x^{2} + 7 x + 14 = A x^{2} - A + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.7

Cancel the common factor of $x + 1$ .

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Step 7.1.10.7.1

Cancel the common factor.

$7 x^{2} + 7 x + 14 = A x^{2} - A + \frac{B (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.7.2

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

$7 x^{2} + 7 x + 14 = A x^{2} - A + (B) (x (x - 1)) + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.8

Apply the distributive property.

$7 x^{2} + 7 x + 14 = A x^{2} - A + B (x \cdot x + x \cdot - 1) + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.9

Multiply $x$ by $x$ .

$7 x^{2} + 7 x + 14 = A x^{2} - A + B (x^{2} + x \cdot - 1) + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.10

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

$7 x^{2} + 7 x + 14 = A x^{2} - A + B (x^{2} - 1 \cdot x) + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.11

Rewrite $- 1 x$ as $- x$ .

$7 x^{2} + 7 x + 14 = A x^{2} - A + B (x^{2} - x) + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.12

Apply the distributive property.

$7 x^{2} + 7 x + 14 = A x^{2} - A + B x^{2} + B (- x) + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.13

Rewrite using the commutative property of multiplication.

$7 x^{2} + 7 x + 14 = A x^{2} - A + B x^{2} - B x + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.14

Cancel the common factor of $x - 1$ .

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Step 7.1.10.14.1

Cancel the common factor.

$7 x^{2} + 7 x + 14 = A x^{2} - A + B x^{2} - B x + \frac{C (x (x + 1) (x - 1))}{x - 1}$

Step 7.1.10.14.2

Divide $(C) (x (x + 1))$ by $1$ .

$7 x^{2} + 7 x + 14 = A x^{2} - A + B x^{2} - B x + (C) (x (x + 1))$

Step 7.1.10.15

Apply the distributive property.

$7 x^{2} + 7 x + 14 = A x^{2} - A + B x^{2} - B x + C (x \cdot x + x \cdot 1)$

Step 7.1.10.16

Multiply $x$ by $x$ .

$7 x^{2} + 7 x + 14 = A x^{2} - A + B x^{2} - B x + C (x^{2} + x \cdot 1)$

Step 7.1.10.17

Multiply $x$ by $1$ .

$7 x^{2} + 7 x + 14 = A x^{2} - A + B x^{2} - B x + C (x^{2} + x)$

Step 7.1.10.18

Apply the distributive property.

$7 x^{2} + 7 x + 14 = A x^{2} - A + B x^{2} - B x + C x^{2} + C x$

Step 7.1.11

Simplify the expression.

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Step 7.1.11.1

Move $B$ .

$7 x^{2} + 7 x + 14 = A x^{2} - A + B x^{2} - x B + C x^{2} + C x$

Step 7.1.11.2

Move $- A$ .

$7 x^{2} + 7 x + 14 = A x^{2} + B x^{2} - x B + C x^{2} + C x - A$

Step 7.1.11.3

Move $- x B$ .

$7 x^{2} + 7 x + 14 = A x^{2} + B x^{2} + C x^{2} - x B + C x - A$

Step 7.2

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

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Step 7.2.1

Create an equation for the partial fraction variables by equating the coefficients of $x^{2}$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$7 = A + B + C$

Step 7.2.2

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.

$7 = - 1 B + C$

Step 7.2.3

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.

$14 = - 1 A$

Step 7.2.4

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

$7 = A + B + C$

$7 = - 1 B + C$

$14 = - 1 A$

$7 = A + B + C$

$7 = - 1 B + C$

$14 = - 1 A$

Step 7.3

Solve the system of equations.

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Step 7.3.1

Solve for $A$ in $14 = - 1 A$ .

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Step 7.3.1.1

Rewrite the equation as $- 1 A = 14$ .

$- 1 A = 14$

$7 = A + B + C$

$7 = - 1 B + C$

Step 7.3.1.2

Divide each term in $- 1 A = 14$ by $- 1$ and simplify.

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Step 7.3.1.2.1

Divide each term in $- 1 A = 14$ by $- 1$ .

$\frac{- 1 A}{- 1} = \frac{14}{- 1}$

$7 = A + B + C$

$7 = - 1 B + C$

Step 7.3.1.2.2

Simplify the left side.

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Step 7.3.1.2.2.1

Dividing two negative values results in a positive value.

$\frac{A}{1} = \frac{14}{- 1}$

$7 = A + B + C$

$7 = - 1 B + C$

Step 7.3.1.2.2.2

Divide $A$ by $1$ .

$A = \frac{14}{- 1}$

$7 = A + B + C$

$7 = - 1 B + C$

$A = \frac{14}{- 1}$

$7 = A + B + C$

$7 = - 1 B + C$

Step 7.3.1.2.3

Simplify the right side.

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Step 7.3.1.2.3.1

Divide $14$ by $- 1$ .

$A = - 14$

$7 = A + B + C$

$7 = - 1 B + C$

$A = - 14$

$7 = A + B + C$

$7 = - 1 B + C$

$A = - 14$

$7 = A + B + C$

$7 = - 1 B + C$

$A = - 14$

$7 = A + B + C$

$7 = - 1 B + C$

Step 7.3.2

Replace all occurrences of $A$ with $- 14$ in each equation.

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Step 7.3.2.1

Replace all occurrences of $A$ in $7 = A + B + C$ with $- 14$ .

$7 = (- 14) + B + C$

$A = - 14$

$7 = - 1 B + C$

Step 7.3.2.2

Simplify the right side.

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Step 7.3.2.2.1

Remove parentheses.

$7 = - 14 + B + C$

$A = - 14$

$7 = - 1 B + C$

$7 = - 14 + B + C$

$A = - 14$

$7 = - 1 B + C$

$7 = - 14 + B + C$

$A = - 14$

$7 = - 1 B + C$

Step 7.3.3

Solve for $B$ in $7 = - 14 + B + C$ .

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Step 7.3.3.1

Rewrite the equation as $- 14 + B + C = 7$ .

$- 14 + B + C = 7$

$A = - 14$

$7 = - 1 B + C$

Step 7.3.3.2

Move all terms not containing $B$ to the right side of the equation.

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Step 7.3.3.2.1

Add $14$ to both sides of the equation.

$B + C = 7 + 14$

$A = - 14$

$7 = - 1 B + C$

Step 7.3.3.2.2

Subtract $C$ from both sides of the equation.

$B = 7 + 14 - C$

$A = - 14$

$7 = - 1 B + C$

Step 7.3.3.2.3

Add $7$ and $14$ .

$B = 21 - C$

$A = - 14$

$7 = - 1 B + C$

$B = 21 - C$

$A = - 14$

$7 = - 1 B + C$

$B = 21 - C$

$A = - 14$

$7 = - 1 B + C$

Step 7.3.4

Replace all occurrences of $B$ with $21 - C$ in each equation.

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Step 7.3.4.1

Replace all occurrences of $B$ in $7 = - 1 B + C$ with $21 - C$ .

$7 = - 1 (21 - C) + C$

$B = 21 - C$

$A = - 14$

Step 7.3.4.2

Simplify the right side.

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Step 7.3.4.2.1

Simplify $- 1 (21 - C) + C$ .

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Step 7.3.4.2.1.1

Simplify each term.

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Step 7.3.4.2.1.1.1

Apply the distributive property.

$7 = - 1 \cdot 21 - 1 (- C) + C$

$B = 21 - C$

$A = - 14$

Step 7.3.4.2.1.1.2

Multiply $- 1$ by $21$ .

$7 = - 21 - 1 (- C) + C$

$B = 21 - C$

$A = - 14$

Step 7.3.4.2.1.1.3

Multiply $- 1 (- C)$ .

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Step 7.3.4.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

$7 = - 21 + 1 C + C$

$B = 21 - C$

$A = - 14$

Step 7.3.4.2.1.1.3.2

Multiply $C$ by $1$ .

$7 = - 21 + C + C$

$B = 21 - C$

$A = - 14$

$7 = - 21 + C + C$

$B = 21 - C$

$A = - 14$

$7 = - 21 + C + C$

$B = 21 - C$

$A = - 14$

Step 7.3.4.2.1.2

Add $C$ and $C$ .

$7 = - 21 + 2 C$

$B = 21 - C$

$A = - 14$

$7 = - 21 + 2 C$

$B = 21 - C$

$A = - 14$

$7 = - 21 + 2 C$

$B = 21 - C$

$A = - 14$

$7 = - 21 + 2 C$

$B = 21 - C$

$A = - 14$

Step 7.3.5

Solve for $C$ in $7 = - 21 + 2 C$ .

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Step 7.3.5.1

Rewrite the equation as $- 21 + 2 C = 7$ .

$- 21 + 2 C = 7$

$B = 21 - C$

$A = - 14$

Step 7.3.5.2

Move all terms not containing $C$ to the right side of the equation.

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Step 7.3.5.2.1

Add $21$ to both sides of the equation.

$2 C = 7 + 21$

$B = 21 - C$

$A = - 14$

Step 7.3.5.2.2

Add $7$ and $21$ .

$2 C = 28$

$B = 21 - C$

$A = - 14$

$2 C = 28$

$B = 21 - C$

$A = - 14$

Step 7.3.5.3

Divide each term in $2 C = 28$ by $2$ and simplify.

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Step 7.3.5.3.1

Divide each term in $2 C = 28$ by $2$ .

$\frac{2 C}{2} = \frac{28}{2}$

$B = 21 - C$

$A = - 14$

Step 7.3.5.3.2

Simplify the left side.

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Step 7.3.5.3.2.1

Cancel the common factor of $2$ .

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Step 7.3.5.3.2.1.1

Cancel the common factor.

$\frac{2 C}{2} = \frac{28}{2}$

$B = 21 - C$

$A = - 14$

Step 7.3.5.3.2.1.2

Divide $C$ by $1$ .

$C = \frac{28}{2}$

$B = 21 - C$

$A = - 14$

$C = \frac{28}{2}$

$B = 21 - C$

$A = - 14$

$C = \frac{28}{2}$

$B = 21 - C$

$A = - 14$

Step 7.3.5.3.3

Simplify the right side.

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Step 7.3.5.3.3.1

Divide $28$ by $2$ .

$C = 14$

$B = 21 - C$

$A = - 14$

$C = 14$

$B = 21 - C$

$A = - 14$

$C = 14$

$B = 21 - C$

$A = - 14$

$C = 14$

$B = 21 - C$

$A = - 14$

Step 7.3.6

Replace all occurrences of $C$ with $14$ in each equation.

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Step 7.3.6.1

Replace all occurrences of $C$ in $B = 21 - C$ with $14$ .

$B = 21 - (14)$

$C = 14$

$A = - 14$

Step 7.3.6.2

Simplify the right side.

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Step 7.3.6.2.1

Simplify $21 - (14)$ .

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Step 7.3.6.2.1.1

Multiply $- 1$ by $14$ .

$B = 21 - 14$

$C = 14$

$A = - 14$

Step 7.3.6.2.1.2

Subtract $14$ from $21$ .

$B = 7$

$C = 14$

$A = - 14$

$B = 7$

$C = 14$

$A = - 14$

$B = 7$

$C = 14$

$A = - 14$

$B = 7$

$C = 14$

$A = - 14$

Step 7.3.7

List all of the solutions.

$B = 7, C = 14, A = - 14$

Step 7.4

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

$- \frac{14}{x} + \frac{7}{x + 1} + \frac{14}{x - 1}$

Step 7.5

Move the negative in front of the fraction.

$7 (\frac{x^{3}}{3} + C) + 7 x + C + \int - \frac{14}{x} + \frac{7}{x + 1} + \frac{14}{x - 1} d x$

Step 8

Split the single integral into multiple integrals.

$7 (\frac{x^{3}}{3} + C) + 7 x + C + \int - \frac{14}{x} d x + \int \frac{7}{x + 1} d x + \int \frac{14}{x - 1} d x$

Step 9

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

$7 (\frac{x^{3}}{3} + C) + 7 x + C - \int \frac{14}{x} d x + \int \frac{7}{x + 1} d x + \int \frac{14}{x - 1} d x$

Step 10

Since $14$ is constant with respect to $x$ , move $14$ out of the integral.

$7 (\frac{x^{3}}{3} + C) + 7 x + C - (14 \int \frac{1}{x} d x) + \int \frac{7}{x + 1} d x + \int \frac{14}{x - 1} d x$

Step 11

Multiply $14$ by $- 1$ .

$7 (\frac{x^{3}}{3} + C) + 7 x + C - 14 \int \frac{1}{x} d x + \int \frac{7}{x + 1} d x + \int \frac{14}{x - 1} d x$

Step 12

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

$7 (\frac{x^{3}}{3} + C) + 7 x + C - 14 (\ln (| x |) + C) + \int \frac{7}{x + 1} d x + \int \frac{14}{x - 1} d x$

Step 13

Since $7$ is constant with respect to $x$ , move $7$ out of the integral.

$7 (\frac{x^{3}}{3} + C) + 7 x + C - 14 (\ln (| x |) + C) + 7 \int \frac{1}{x + 1} d x + \int \frac{14}{x - 1} d x$

Step 14

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

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Step 14.1

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

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Step 14.1.1

Differentiate $x + 1$ .

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

Step 14.1.2

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 14.1.3

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 14.1.4

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$1 + 0$

Step 14.1.5

Add $1$ and $0$ .

$1$

Step 14.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$7 (\frac{x^{3}}{3} + C) + 7 x + C - 14 (\ln (| x |) + C) + 7 \int \frac{1}{u_{1}} d u_{1} + \int \frac{14}{x - 1} d x$

Step 15

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

$7 (\frac{x^{3}}{3} + C) + 7 x + C - 14 (\ln (| x |) + C) + 7 (\ln (| u_{1} |) + C) + \int \frac{14}{x - 1} d x$

Step 16

Since $14$ is constant with respect to $x$ , move $14$ out of the integral.

$7 (\frac{x^{3}}{3} + C) + 7 x + C - 14 (\ln (| x |) + C) + 7 (\ln (| u_{1} |) + C) + 14 \int \frac{1}{x - 1} d x$

Step 17

Let $u_{2} = x - 1$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 17.1

Let $u_{2} = x - 1$ . Find $\frac{d u_{2}}{d x}$ .

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Step 17.1.1

Differentiate $x - 1$ .

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

Step 17.1.2

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 17.1.3

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 17.1.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$1 + 0$

Step 17.1.5

Add $1$ and $0$ .

$1$

Step 17.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$7 (\frac{x^{3}}{3} + C) + 7 x + C - 14 (\ln (| x |) + C) + 7 (\ln (| u_{1} |) + C) + 14 \int \frac{1}{u_{2}} d u_{2}$

Step 18

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

$7 (\frac{x^{3}}{3} + C) + 7 x + C - 14 (\ln (| x |) + C) + 7 (\ln (| u_{1} |) + C) + 14 (\ln (| u_{2} |) + C)$

Step 19

Simplify.

$\frac{7 x^{3}}{3} + 7 x - 14 \ln (| x |) + 7 \ln (| u_{1} |) + 14 \ln (| u_{2} |) + C$

Step 20

Substitute back in for each integration substitution variable.

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Step 20.1

Replace all occurrences of $u_{1}$ with $x + 1$ .

$\frac{7 x^{3}}{3} + 7 x - 14 \ln (| x |) + 7 \ln (| x + 1 |) + 14 \ln (| u_{2} |) + C$

Step 20.2

Replace all occurrences of $u_{2}$ with $x - 1$ .

$\frac{7 x^{3}}{3} + 7 x - 14 \ln (| x |) + 7 \ln (| x + 1 |) + 14 \ln (| x - 1 |) + C$

Step 21

Reorder terms.

$\frac{7}{3} x^{3} + 7 x - 14 \ln (| x |) + 7 \ln (| x + 1 |) + 14 \ln (| x - 1 |) + C$