Calculus Examples

$\int \frac{2 x^{2} - 5 x - 2}{{(x - 2)}^{2} (x - 1)} d x$

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

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 $A$ .

$\frac{A}{{(x - 2)}^{2}}$

Step 1.1.2

$\frac{A}{{(x - 2)}^{2}} + \frac{B}{x - 2}$

Step 1.1.3

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

Step 1.1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is ${(x - 2)}^{2} (x - 1)$ .

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

Step 1.1.5

Cancel the common factor of ${(x - 2)}^{2}$ .

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Step 1.1.5.1

Cancel the common factor.

Step 1.1.5.2

Rewrite the expression.

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

Step 1.1.6

Cancel the common factor of $x - 1$ .

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Step 1.1.6.1

Cancel the common factor.

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

Step 1.1.6.2

Divide $2 x^{2} - 5 x - 2$ by $1$ .

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

Step 1.1.7

Simplify each term.

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Step 1.1.7.1

Cancel the common factor of ${(x - 2)}^{2}$ .

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Step 1.1.7.1.1

Cancel the common factor.

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

Step 1.1.7.1.2

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

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

Step 1.1.7.2

Apply the distributive property.

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

Step 1.1.7.3

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

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

Step 1.1.7.4

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

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

Step 1.1.7.5

Cancel the common factor of ${(x - 2)}^{2}$ and $x - 2$ .

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Step 1.1.7.5.1

Factor $x - 2$ out of $(B) {(x - 2)}^{2} (x - 1)$ .

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

Step 1.1.7.5.2

Cancel the common factors.

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Step 1.1.7.5.2.1

Multiply by $1$ .

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

Step 1.1.7.5.2.2

Cancel the common factor.

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

Step 1.1.7.5.2.3

Rewrite the expression.

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

Step 1.1.7.5.2.4

Divide $B (x - 2) (x - 1)$ by $1$ .

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

Step 1.1.7.6

Apply the distributive property.

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

Step 1.1.7.7

Move $- 2$ to the left of $B$ .

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

Step 1.1.7.8

Expand $(B x - 2 B) (x - 1)$ using the FOIL Method.

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Step 1.1.7.8.1

Apply the distributive property.

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

Step 1.1.7.8.2

Apply the distributive property.

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

Step 1.1.7.8.3

Apply the distributive property.

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

Step 1.1.7.9

Simplify and combine like terms.

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Step 1.1.7.9.1

Simplify each term.

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Step 1.1.7.9.1.1

Multiply $x$ by $x$ by adding the exponents.

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Step 1.1.7.9.1.1.1

Move $x$ .

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

Step 1.1.7.9.1.1.2

Multiply $x$ by $x$ .

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

Step 1.1.7.9.1.2

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

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

Step 1.1.7.9.1.3

Rewrite $- 1 (B x)$ as $- (B x)$ .

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

Step 1.1.7.9.1.4

Multiply $- 1$ by $- 2$ .

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

Step 1.1.7.9.2

Subtract $2 B x$ from $- B x$ .

$2 x^{2} - 5 x - 2 = A x - A + B x^{2} - 3 B x + 2 B + \frac{(C) {(x - 2)}^{2} (x - 1)}{x - 1}$

Step 1.1.7.10

Cancel the common factor of $x - 1$ .

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Step 1.1.7.10.1

Cancel the common factor.

$2 x^{2} - 5 x - 2 = A x - A + B x^{2} - 3 B x + 2 B + \frac{(C) {(x - 2)}^{2} (x - 1)}{x - 1}$

Step 1.1.7.10.2

Divide $(C) {(x - 2)}^{2}$ by $1$ .

$2 x^{2} - 5 x - 2 = A x - A + B x^{2} - 3 B x + 2 B + (C) {(x - 2)}^{2}$

Step 1.1.7.11

Rewrite ${(x - 2)}^{2}$ as $(x - 2) (x - 2)$ .

$2 x^{2} - 5 x - 2 = A x - A + B x^{2} - 3 B x + 2 B + C ((x - 2) (x - 2))$

Step 1.1.7.12

Expand $(x - 2) (x - 2)$ using the FOIL Method.

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Step 1.1.7.12.1

Apply the distributive property.

$2 x^{2} - 5 x - 2 = A x - A + B x^{2} - 3 B x + 2 B + C (x (x - 2) - 2 (x - 2))$

Step 1.1.7.12.2

Apply the distributive property.

$2 x^{2} - 5 x - 2 = A x - A + B x^{2} - 3 B x + 2 B + C (x \cdot x + x \cdot - 2 - 2 (x - 2))$

Step 1.1.7.12.3

Apply the distributive property.

$2 x^{2} - 5 x - 2 = A x - A + B x^{2} - 3 B x + 2 B + C (x \cdot x + x \cdot - 2 - 2 x - 2 \cdot - 2)$

Step 1.1.7.13

Simplify and combine like terms.

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Step 1.1.7.13.1

Simplify each term.

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Step 1.1.7.13.1.1

Multiply $x$ by $x$ .

$2 x^{2} - 5 x - 2 = A x - A + B x^{2} - 3 B x + 2 B + C (x^{2} + x \cdot - 2 - 2 x - 2 \cdot - 2)$

Step 1.1.7.13.1.2

Move $- 2$ to the left of $x$ .

$2 x^{2} - 5 x - 2 = A x - A + B x^{2} - 3 B x + 2 B + C (x^{2} - 2 \cdot x - 2 x - 2 \cdot - 2)$

Step 1.1.7.13.1.3

Multiply $- 2$ by $- 2$ .

$2 x^{2} - 5 x - 2 = A x - A + B x^{2} - 3 B x + 2 B + C (x^{2} - 2 x - 2 x + 4)$

Step 1.1.7.13.2

Subtract $2 x$ from $- 2 x$ .

$2 x^{2} - 5 x - 2 = A x - A + B x^{2} - 3 B x + 2 B + C (x^{2} - 4 x + 4)$

Step 1.1.7.14

Apply the distributive property.

$2 x^{2} - 5 x - 2 = A x - A + B x^{2} - 3 B x + 2 B + C x^{2} + C (- 4 x) + C \cdot 4$

Step 1.1.7.15

Simplify.

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Step 1.1.7.15.1

Rewrite using the commutative property of multiplication.

$2 x^{2} - 5 x - 2 = A x - A + B x^{2} - 3 B x + 2 B + C x^{2} - 4 C x + C \cdot 4$

Step 1.1.7.15.2

Move $4$ to the left of $C$ .

$2 x^{2} - 5 x - 2 = A x - A + B x^{2} - 3 B x + 2 B + C x^{2} - 4 C x + 4 \cdot C$

$2 x^{2} - 5 x - 2 = A x - A + B x^{2} - 3 B x + 2 B + C x^{2} - 4 C x + 4 C$

Step 1.1.8

Simplify the expression.

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Step 1.1.8.1

Move $B$ .

$2 x^{2} - 5 x - 2 = A x - A + B x^{2} - 3 x B + 2 B + C x^{2} - 4 C x + 4 C$

Step 1.1.8.2

Move $2 B$ .

$2 x^{2} - 5 x - 2 = A x - A + B x^{2} - 3 x B + C x^{2} - 4 C x + 2 B + 4 C$

Step 1.1.8.3

Move $- A$ .

$2 x^{2} - 5 x - 2 = A x + B x^{2} - 3 x B + C x^{2} - 4 C x - A + 2 B + 4 C$

Step 1.1.8.4

Move $- 3 x B$ .

$2 x^{2} - 5 x - 2 = A x + B x^{2} + C x^{2} - 3 x B - 4 C x - A + 2 B + 4 C$

Step 1.1.8.5

Move $A x$ .

$2 x^{2} - 5 x - 2 = B x^{2} + C x^{2} + A x - 3 x B - 4 C x - A + 2 B + 4 C$

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 $x^{2}$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$2 = B + C$

Step 1.2.2

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.

$- 5 = A - 3 B - 4 C$

Step 1.2.3

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.

$- 2 = - 1 A + 2 B + 4 C$

Step 1.2.4

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

$2 = B + C$

$- 5 = A - 3 B - 4 C$

$- 2 = - 1 A + 2 B + 4 C$

$2 = B + C$

$- 5 = A - 3 B - 4 C$

$- 2 = - 1 A + 2 B + 4 C$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Solve for $B$ in $2 = B + C$ .

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Step 1.3.1.1

Rewrite the equation as $B + C = 2$ .

$B + C = 2$

$- 5 = A - 3 B - 4 C$

$- 2 = - 1 A + 2 B + 4 C$

Step 1.3.1.2

Subtract $C$ from both sides of the equation.

$B = 2 - C$

$- 5 = A - 3 B - 4 C$

$- 2 = - 1 A + 2 B + 4 C$

$B = 2 - C$

$- 5 = A - 3 B - 4 C$

$- 2 = - 1 A + 2 B + 4 C$

Step 1.3.2

Replace all occurrences of $B$ with $2 - C$ in each equation.

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Step 1.3.2.1

Replace all occurrences of $B$ in $- 5 = A - 3 B - 4 C$ with $2 - C$ .

$- 5 = A - 3 (2 - C) - 4 C$

$B = 2 - C$

$- 2 = - 1 A + 2 B + 4 C$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Simplify $A - 3 (2 - C) - 4 C$ .

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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.

$- 5 = A - 3 \cdot 2 - 3 (- C) - 4 C$

$B = 2 - C$

$- 2 = - 1 A + 2 B + 4 C$

Step 1.3.2.2.1.1.2

Multiply $- 3$ by $2$ .

$- 5 = A - 6 - 3 (- C) - 4 C$

$B = 2 - C$

$- 2 = - 1 A + 2 B + 4 C$

Step 1.3.2.2.1.1.3

Multiply $- 1$ by $- 3$ .

$- 5 = A - 6 + 3 C - 4 C$

$B = 2 - C$

$- 2 = - 1 A + 2 B + 4 C$

$- 5 = A - 6 + 3 C - 4 C$

$B = 2 - C$

$- 2 = - 1 A + 2 B + 4 C$

Step 1.3.2.2.1.2

Subtract $4 C$ from $3 C$ .

$- 5 = A - 6 - C$

$B = 2 - C$

$- 2 = - 1 A + 2 B + 4 C$

$- 5 = A - 6 - C$

$B = 2 - C$

$- 2 = - 1 A + 2 B + 4 C$

$- 5 = A - 6 - C$

$B = 2 - C$

$- 2 = - 1 A + 2 B + 4 C$

Step 1.3.2.3

Replace all occurrences of $B$ in $- 2 = - 1 A + 2 B + 4 C$ with $2 - C$ .

$- 2 = - 1 A + 2 (2 - C) + 4 C$

$- 5 = A - 6 - C$

$B = 2 - C$

Step 1.3.2.4

Simplify the right side.

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Step 1.3.2.4.1

Simplify $- 1 A + 2 (2 - C) + 4 C$ .

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Step 1.3.2.4.1.1

Simplify each term.

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Step 1.3.2.4.1.1.1

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

$- 2 = - A + 2 (2 - C) + 4 C$

$- 5 = A - 6 - C$

$B = 2 - C$

Step 1.3.2.4.1.1.2

Apply the distributive property.

$- 2 = - A + 2 \cdot 2 + 2 (- C) + 4 C$

$- 5 = A - 6 - C$

$B = 2 - C$

Step 1.3.2.4.1.1.3

Multiply $2$ by $2$ .

$- 2 = - A + 4 + 2 (- C) + 4 C$

$- 5 = A - 6 - C$

$B = 2 - C$

Step 1.3.2.4.1.1.4

Multiply $- 1$ by $2$ .

$- 2 = - A + 4 - 2 C + 4 C$

$- 5 = A - 6 - C$

$B = 2 - C$

$- 2 = - A + 4 - 2 C + 4 C$

$- 5 = A - 6 - C$

$B = 2 - C$

Step 1.3.2.4.1.2

Add $- 2 C$ and $4 C$ .

$- 2 = - A + 4 + 2 C$

$- 5 = A - 6 - C$

$B = 2 - C$

$- 2 = - A + 4 + 2 C$

$- 5 = A - 6 - C$

$B = 2 - C$

$- 2 = - A + 4 + 2 C$

$- 5 = A - 6 - C$

$B = 2 - C$

$- 2 = - A + 4 + 2 C$

$- 5 = A - 6 - C$

$B = 2 - C$

Step 1.3.3

Reorder $2$ and $- C$ .

$B = - C + 2$

$- 2 = - A + 4 + 2 C$

$- 5 = A - 6 - C$

Step 1.3.4

Solve for $A$ in $- 5 = A - 6 - C$ .

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Step 1.3.4.1

Rewrite the equation as $A - 6 - C = - 5$ .

$A - 6 - C = - 5$

$B = - C + 2$

$- 2 = - A + 4 + 2 C$

Step 1.3.4.2

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

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Step 1.3.4.2.1

Add $6$ to both sides of the equation.

$A - C = - 5 + 6$

$B = - C + 2$

$- 2 = - A + 4 + 2 C$

Step 1.3.4.2.2

Add $C$ to both sides of the equation.

$A = - 5 + 6 + C$

$B = - C + 2$

$- 2 = - A + 4 + 2 C$

Step 1.3.4.2.3

Add $- 5$ and $6$ .

$A = 1 + C$

$B = - C + 2$

$- 2 = - A + 4 + 2 C$

$A = 1 + C$

$B = - C + 2$

$- 2 = - A + 4 + 2 C$

$A = 1 + C$

$B = - C + 2$

$- 2 = - A + 4 + 2 C$

Step 1.3.5

Replace all occurrences of $A$ with $1 + C$ in each equation.

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Step 1.3.5.1

Replace all occurrences of $A$ in $- 2 = - A + 4 + 2 C$ with $1 + C$ .

$- 2 = - (1 + C) + 4 + 2 C$

$A = 1 + C$

$B = - C + 2$

Step 1.3.5.2

Simplify the right side.

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Step 1.3.5.2.1

Simplify $- (1 + C) + 4 + 2 C$ .

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Step 1.3.5.2.1.1

Simplify each term.

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Step 1.3.5.2.1.1.1

Apply the distributive property.

$- 2 = - 1 \cdot 1 - C + 4 + 2 C$

$A = 1 + C$

$B = - C + 2$

Step 1.3.5.2.1.1.2

Multiply $- 1$ by $1$ .

$- 2 = - 1 - C + 4 + 2 C$

$A = 1 + C$

$B = - C + 2$

$- 2 = - 1 - C + 4 + 2 C$

$A = 1 + C$

$B = - C + 2$

Step 1.3.5.2.1.2

Simplify by adding terms.

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Step 1.3.5.2.1.2.1

Add $- 1$ and $4$ .

$- 2 = - C + 3 + 2 C$

$A = 1 + C$

$B = - C + 2$

Step 1.3.5.2.1.2.2

Add $- C$ and $2 C$ .

$- 2 = C + 3$

$A = 1 + C$

$B = - C + 2$

$- 2 = C + 3$

$A = 1 + C$

$B = - C + 2$

$- 2 = C + 3$

$A = 1 + C$

$B = - C + 2$

$- 2 = C + 3$

$A = 1 + C$

$B = - C + 2$

$- 2 = C + 3$

$A = 1 + C$

$B = - C + 2$

Step 1.3.6

Solve for $C$ in $- 2 = C + 3$ .

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Step 1.3.6.1

Rewrite the equation as $C + 3 = - 2$ .

$C + 3 = - 2$

$A = 1 + C$

$B = - C + 2$

Step 1.3.6.2

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

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Step 1.3.6.2.1

Subtract $3$ from both sides of the equation.

$C = - 2 - 3$

$A = 1 + C$

$B = - C + 2$

Step 1.3.6.2.2

Subtract $3$ from $- 2$ .

$C = - 5$

$A = 1 + C$

$B = - C + 2$

$C = - 5$

$A = 1 + C$

$B = - C + 2$

$C = - 5$

$A = 1 + C$

$B = - C + 2$

Step 1.3.7

Replace all occurrences of $C$ with $- 5$ in each equation.

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Step 1.3.7.1

Replace all occurrences of $C$ in $A = 1 + C$ with $- 5$ .

$A = 1 - 5$

$C = - 5$

$B = - C + 2$

Step 1.3.7.2

Simplify $A = 1 - 5$ .

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Step 1.3.7.2.1

Simplify the left side.

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Step 1.3.7.2.1.1

Remove parentheses.

$A = 1 - 5$

$C = - 5$

$B = - C + 2$

$A = 1 - 5$

$C = - 5$

$B = - C + 2$

Step 1.3.7.2.2

Simplify the right side.

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Step 1.3.7.2.2.1

Subtract $5$ from $1$ .

$A = - 4$

$C = - 5$

$B = - C + 2$

$A = - 4$

$C = - 5$

$B = - C + 2$

$A = - 4$

$C = - 5$

$B = - C + 2$

Step 1.3.7.3

Replace all occurrences of $C$ in $B = - C + 2$ with $- 5$ .

$B = - (- 5) + 2$

$A = - 4$

$C = - 5$

Step 1.3.7.4

Simplify the right side.

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Step 1.3.7.4.1

Simplify $- (- 5) + 2$ .

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Step 1.3.7.4.1.1

Multiply $- 1$ by $- 5$ .

$B = 5 + 2$

$A = - 4$

$C = - 5$

Step 1.3.7.4.1.2

Add $5$ and $2$ .

$B = 7$

$A = - 4$

$C = - 5$

$B = 7$

$A = - 4$

$C = - 5$

$B = 7$

$A = - 4$

$C = - 5$

$B = 7$

$A = - 4$

$C = - 5$

Step 1.3.8

List all of the solutions.

$B = 7, A = - 4, C = - 5$

Step 1.4

Replace each of the partial fraction coefficients in $\frac{A}{{(x - 2)}^{2}} + \frac{B}{x - 2} + \frac{C}{x - 1}$ with the values found for $A$ , $B$ , and $C$ .

$\frac{- 4}{{(x - 2)}^{2}} + \frac{7}{x - 2} + \frac{- 5}{x - 1}$

Step 1.5

Simplify.

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Step 1.5.1

Move the negative in front of the fraction.

$- \frac{4}{{(x - 2)}^{2}} + \frac{7}{x - 2} + \frac{- 5}{x - 1}$

Step 1.5.2

Move the negative in front of the fraction.

$\int - \frac{4}{{(x - 2)}^{2}} + \frac{7}{x - 2} - \frac{5}{x - 1} d x$

Step 2

Split the single integral into multiple integrals.

$\int - \frac{4}{{(x - 2)}^{2}} d x + \int \frac{7}{x - 2} d x + \int - \frac{5}{x - 1} d x$

Step 3

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

$- \int \frac{4}{{(x - 2)}^{2}} d x + \int \frac{7}{x - 2} d x + \int - \frac{5}{x - 1} d x$

Step 4

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

$- (4 \int \frac{1}{{(x - 2)}^{2}} d x) + \int \frac{7}{x - 2} d x + \int - \frac{5}{x - 1} d x$

Step 5

Multiply $4$ by $- 1$ .

$- 4 \int \frac{1}{{(x - 2)}^{2}} d x + \int \frac{7}{x - 2} d x + \int - \frac{5}{x - 1} d x$

Step 6

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

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Step 6.1

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

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Step 6.1.1

Differentiate $x - 2$ .

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

Step 6.1.2

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 6.1.3

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

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

Step 6.1.4

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$1 + 0$

Step 6.1.5

Add $1$ and $0$ .

$1$

Step 6.2

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

$- 4 \int \frac{1}{{u_{1}}^{2}} d u_{1} + \int \frac{7}{x - 2} d x + \int - \frac{5}{x - 1} d x$

Step 7

Apply basic rules of exponents.

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Step 7.1

Move ${u_{1}}^{2}$ out of the denominator by raising it to the $- 1$ power.

$- 4 \int {({u_{1}}^{2})}^{- 1} d u_{1} + \int \frac{7}{x - 2} d x + \int - \frac{5}{x - 1} d x$

Step 7.2

Multiply the exponents in ${({u_{1}}^{2})}^{- 1}$ .

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Step 7.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$- 4 \int {u_{1}}^{2 \cdot - 1} d u_{1} + \int \frac{7}{x - 2} d x + \int - \frac{5}{x - 1} d x$

Step 7.2.2

Multiply $2$ by $- 1$ .

$- 4 \int {u_{1}}^{- 2} d u_{1} + \int \frac{7}{x - 2} d x + \int - \frac{5}{x - 1} d x$

Step 8

By the Power Rule, the integral of ${u_{1}}^{- 2}$ with respect to $u_{1}$ is $- {u_{1}}^{- 1}$ .

$- 4 (- {u_{1}}^{- 1} + C) + \int \frac{7}{x - 2} d x + \int - \frac{5}{x - 1} d x$

Step 9

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

$- 4 (- {u_{1}}^{- 1} + C) + 7 \int \frac{1}{x - 2} d x + \int - \frac{5}{x - 1} d x$

Step 10

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

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Step 10.1

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

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Step 10.1.1

Differentiate $x - 2$ .

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

Step 10.1.2

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 10.1.3

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 10.1.4

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$1 + 0$

Step 10.1.5

Add $1$ and $0$ .

$1$

Step 10.2

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

$- 4 (- {u_{1}}^{- 1} + C) + 7 \int \frac{1}{u_{2}} d u_{2} + \int - \frac{5}{x - 1} d x$

Step 11

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

$- 4 (- {u_{1}}^{- 1} + C) + 7 (\ln (| u_{2} |) + C) + \int - \frac{5}{x - 1} d x$

Step 12

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

$- 4 (- {u_{1}}^{- 1} + C) + 7 (\ln (| u_{2} |) + C) - \int \frac{5}{x - 1} d x$

Step 13

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

$- 4 (- {u_{1}}^{- 1} + C) + 7 (\ln (| u_{2} |) + C) - (5 \int \frac{1}{x - 1} d x)$

Step 14

Multiply $5$ by $- 1$ .

$- 4 (- {u_{1}}^{- 1} + C) + 7 (\ln (| u_{2} |) + C) - 5 \int \frac{1}{x - 1} d x$

Step 15

Let $u_{3} = x - 1$ . Then $d u_{3} = d x$ . Rewrite using $u_{3}$ and $d$ $u_{3}$ .

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Step 15.1

Let $u_{3} = x - 1$ . Find $\frac{d u_{3}}{d x}$ .

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Step 15.1.1

Differentiate $x - 1$ .

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

Step 15.1.2

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 15.1.3

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 15.1.4

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

$1 + 0$

Step 15.1.5

Add $1$ and $0$ .

$1$

Step 15.2

Rewrite the problem using $u_{3}$ and $d u_{3}$ .

$- 4 (- {u_{1}}^{- 1} + C) + 7 (\ln (| u_{2} |) + C) - 5 \int \frac{1}{u_{3}} d u_{3}$

Step 16

The integral of $\frac{1}{u_{3}}$ with respect to $u_{3}$ is $\ln (| u_{3} |)$ .

$- 4 (- {u_{1}}^{- 1} + C) + 7 (\ln (| u_{2} |) + C) - 5 (\ln (| u_{3} |) + C)$

Step 17

Simplify.

$\frac{4}{u_{1}} + 7 \ln (| u_{2} |) - 5 \ln (| u_{3} |) + C$

Step 18

Substitute back in for each integration substitution variable.

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Step 18.1

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

$\frac{4}{x - 2} + 7 \ln (| u_{2} |) - 5 \ln (| u_{3} |) + C$

Step 18.2

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

$\frac{4}{x - 2} + 7 \ln (| x - 2 |) - 5 \ln (| u_{3} |) + C$

Step 18.3

Replace all occurrences of $u_{3}$ with $x - 1$ .

$\frac{4}{x - 2} + 7 \ln (| x - 2 |) - 5 \ln (| x - 1 |) + C$