Calculus Examples

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

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 $A$ .

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

Step 1.1.2

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

Step 1.1.3

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

Step 1.1.4

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

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

Step 1.1.5

Cancel the common factor of $z - 1$ .

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Step 1.1.5.1

Cancel the common factor.

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

Step 1.1.5.2

Rewrite the expression.

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

Step 1.1.6

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

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Step 1.1.6.1

Cancel the common factor.

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

Step 1.1.6.2

Divide $z$ by $1$ .

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

Step 1.1.7

Simplify each term.

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Step 1.1.7.1

Cancel the common factor of $z - 1$ .

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Step 1.1.7.1.1

Cancel the common factor.

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

Step 1.1.7.1.2

Divide $(A) {(z - 2)}^{2}$ by $1$ .

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

Step 1.1.7.2

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

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

Step 1.1.7.3

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

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Step 1.1.7.3.1

Apply the distributive property.

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

Step 1.1.7.3.2

Apply the distributive property.

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

Step 1.1.7.3.3

Apply the distributive property.

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

Step 1.1.7.4

Simplify and combine like terms.

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Step 1.1.7.4.1

Simplify each term.

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Step 1.1.7.4.1.1

Multiply $z$ by $z$ .

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

Step 1.1.7.4.1.2

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

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

Step 1.1.7.4.1.3

Multiply $- 2$ by $- 2$ .

$z = A (z^{2} - 2 z - 2 z + 4) + \frac{(B) (z - 1) {(z - 2)}^{2}}{{(z - 2)}^{2}} + \frac{(C) (z - 1) {(z - 2)}^{2}}{z - 2}$

Step 1.1.7.4.2

Subtract $2 z$ from $- 2 z$ .

$z = A (z^{2} - 4 z + 4) + \frac{(B) (z - 1) {(z - 2)}^{2}}{{(z - 2)}^{2}} + \frac{(C) (z - 1) {(z - 2)}^{2}}{z - 2}$

Step 1.1.7.5

Apply the distributive property.

$z = A z^{2} + A (- 4 z) + A \cdot 4 + \frac{(B) (z - 1) {(z - 2)}^{2}}{{(z - 2)}^{2}} + \frac{(C) (z - 1) {(z - 2)}^{2}}{z - 2}$

Step 1.1.7.6

Simplify.

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Step 1.1.7.6.1

Rewrite using the commutative property of multiplication.

$z = A z^{2} - 4 A z + A \cdot 4 + \frac{(B) (z - 1) {(z - 2)}^{2}}{{(z - 2)}^{2}} + \frac{(C) (z - 1) {(z - 2)}^{2}}{z - 2}$

Step 1.1.7.6.2

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

$z = A z^{2} - 4 A z + 4 \cdot A + \frac{(B) (z - 1) {(z - 2)}^{2}}{{(z - 2)}^{2}} + \frac{(C) (z - 1) {(z - 2)}^{2}}{z - 2}$

Step 1.1.7.7

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

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Step 1.1.7.7.1

Cancel the common factor.

$z = A z^{2} - 4 A z + 4 A + \frac{(B) (z - 1) {(z - 2)}^{2}}{{(z - 2)}^{2}} + \frac{(C) (z - 1) {(z - 2)}^{2}}{z - 2}$

Step 1.1.7.7.2

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

$z = A z^{2} - 4 A z + 4 A + (B) (z - 1) + \frac{(C) (z - 1) {(z - 2)}^{2}}{z - 2}$

Step 1.1.7.8

Apply the distributive property.

$z = A z^{2} - 4 A z + 4 A + B z + B \cdot - 1 + \frac{(C) (z - 1) {(z - 2)}^{2}}{z - 2}$

Step 1.1.7.9

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

$z = A z^{2} - 4 A z + 4 A + B z - 1 \cdot B + \frac{(C) (z - 1) {(z - 2)}^{2}}{z - 2}$

Step 1.1.7.10

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

$z = A z^{2} - 4 A z + 4 A + B z - B + \frac{(C) (z - 1) {(z - 2)}^{2}}{z - 2}$

Step 1.1.7.11

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

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Step 1.1.7.11.1

Factor $z - 2$ out of $(C) (z - 1) {(z - 2)}^{2}$ .

$z = A z^{2} - 4 A z + 4 A + B z - B + \frac{(z - 2) ((C) (z - 1) (z - 2))}{z - 2}$

Step 1.1.7.11.2

Cancel the common factors.

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Step 1.1.7.11.2.1

Multiply by $1$ .

$z = A z^{2} - 4 A z + 4 A + B z - B + \frac{(z - 2) ((C) (z - 1) (z - 2))}{(z - 2) \cdot 1}$

Step 1.1.7.11.2.2

Cancel the common factor.

$z = A z^{2} - 4 A z + 4 A + B z - B + \frac{(z - 2) ((C) (z - 1) (z - 2))}{(z - 2) \cdot 1}$

Step 1.1.7.11.2.3

Rewrite the expression.

$z = A z^{2} - 4 A z + 4 A + B z - B + \frac{(C) (z - 1) (z - 2)}{1}$

Step 1.1.7.11.2.4

Divide $(C) (z - 1) (z - 2)$ by $1$ .

$z = A z^{2} - 4 A z + 4 A + B z - B + (C) (z - 1) (z - 2)$

Step 1.1.7.12

Apply the distributive property.

$z = A z^{2} - 4 A z + 4 A + B z - B + (C z + C \cdot - 1) (z - 2)$

Step 1.1.7.13

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

$z = A z^{2} - 4 A z + 4 A + B z - B + (C z - 1 \cdot C) (z - 2)$

Step 1.1.7.14

Rewrite $- 1 C$ as $- C$ .

$z = A z^{2} - 4 A z + 4 A + B z - B + (C z - C) (z - 2)$

Step 1.1.7.15

Expand $(C z - C) (z - 2)$ using the FOIL Method.

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Step 1.1.7.15.1

Apply the distributive property.

$z = A z^{2} - 4 A z + 4 A + B z - B + C z (z - 2) - C (z - 2)$

Step 1.1.7.15.2

Apply the distributive property.

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

Step 1.1.7.15.3

Apply the distributive property.

$z = A z^{2} - 4 A z + 4 A + B z - B + C z \cdot z + C z \cdot - 2 - C z - C \cdot - 2$

Step 1.1.7.16

Simplify and combine like terms.

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Step 1.1.7.16.1

Simplify each term.

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Step 1.1.7.16.1.1

Multiply $z$ by $z$ by adding the exponents.

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Step 1.1.7.16.1.1.1

Move $z$ .

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

Step 1.1.7.16.1.1.2

Multiply $z$ by $z$ .

$z = A z^{2} - 4 A z + 4 A + B z - B + C z^{2} + C z \cdot - 2 - C z - C \cdot - 2$

Step 1.1.7.16.1.2

Move $- 2$ to the left of $C z$ .

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

Step 1.1.7.16.1.3

Multiply $- 2$ by $- 1$ .

$z = A z^{2} - 4 A z + 4 A + B z - B + C z^{2} - 2 C z - C z + 2 C$

Step 1.1.7.16.2

Subtract $C z$ from $- 2 C z$ .

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

Step 1.1.8

Simplify the expression.

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Step 1.1.8.1

Move $A$ .

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

Step 1.1.8.2

Reorder $B$ and $z$ .

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

Step 1.1.8.3

Move $- B$ .

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

Step 1.1.8.4

Move $4 A$ .

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

Step 1.1.8.5

Move $B z$ .

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

Step 1.1.8.6

Move $- 4 z A$ .

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

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

$0 = A + C$

Step 1.2.2

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

$1 = - 4 A + B - 3 C$

Step 1.2.3

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing $z$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = 4 A - 1 B + 2 C$

Step 1.2.4

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

$0 = A + C$

$1 = - 4 A + B - 3 C$

$0 = 4 A - 1 B + 2 C$

$0 = A + C$

$1 = - 4 A + B - 3 C$

$0 = 4 A - 1 B + 2 C$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Solve for $A$ in $0 = A + C$ .

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Step 1.3.1.1

Rewrite the equation as $A + C = 0$ .

$A + C = 0$

$1 = - 4 A + B - 3 C$

$0 = 4 A - 1 B + 2 C$

Step 1.3.1.2

Subtract $C$ from both sides of the equation.

$A = - C$

$1 = - 4 A + B - 3 C$

$0 = 4 A - 1 B + 2 C$

$A = - C$

$1 = - 4 A + B - 3 C$

$0 = 4 A - 1 B + 2 C$

Step 1.3.2

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

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Step 1.3.2.1

Replace all occurrences of $A$ in $1 = - 4 A + B - 3 C$ with $- C$ .

$1 = - 4 (- C) + B - 3 C$

$A = - C$

$0 = 4 A - 1 B + 2 C$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Simplify $- 4 (- C) + B - 3 C$ .

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Step 1.3.2.2.1.1

Multiply $- 1$ by $- 4$ .

$1 = 4 C + B - 3 C$

$A = - C$

$0 = 4 A - 1 B + 2 C$

Step 1.3.2.2.1.2

Subtract $3 C$ from $4 C$ .

$1 = B + C$

$A = - C$

$0 = 4 A - 1 B + 2 C$

$1 = B + C$

$A = - C$

$0 = 4 A - 1 B + 2 C$

$1 = B + C$

$A = - C$

$0 = 4 A - 1 B + 2 C$

Step 1.3.2.3

Replace all occurrences of $A$ in $0 = 4 A - 1 B + 2 C$ with $- C$ .

$0 = 4 (- C) - 1 B + 2 C$

$1 = B + C$

$A = - C$

Step 1.3.2.4

Simplify the right side.

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Step 1.3.2.4.1

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

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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

Multiply $- 1$ by $4$ .

$0 = - 4 C - 1 B + 2 C$

$1 = B + C$

$A = - C$

Step 1.3.2.4.1.1.2

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

$0 = - 4 C - B + 2 C$

$1 = B + C$

$A = - C$

$0 = - 4 C - B + 2 C$

$1 = B + C$

$A = - C$

Step 1.3.2.4.1.2

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

$0 = - B - 2 C$

$1 = B + C$

$A = - C$

$0 = - B - 2 C$

$1 = B + C$

$A = - C$

$0 = - B - 2 C$

$1 = B + C$

$A = - C$

$0 = - B - 2 C$

$1 = B + C$

$A = - C$

Step 1.3.3

Solve for $B$ in $1 = B + C$ .

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Step 1.3.3.1

Rewrite the equation as $B + C = 1$ .

$B + C = 1$

$0 = - B - 2 C$

$A = - C$

Step 1.3.3.2

Subtract $C$ from both sides of the equation.

$B = 1 - C$

$0 = - B - 2 C$

$A = - C$

$B = 1 - C$

$0 = - B - 2 C$

$A = - C$

Step 1.3.4

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

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Step 1.3.4.1

Replace all occurrences of $B$ in $0 = - B - 2 C$ with $1 - C$ .

$0 = - (1 - C) - 2 C$

$B = 1 - C$

$A = - C$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

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

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Step 1.3.4.2.1.1

Simplify each term.

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Step 1.3.4.2.1.1.1

Apply the distributive property.

$0 = - 1 \cdot 1 + C - 2 C$

$B = 1 - C$

$A = - C$

Step 1.3.4.2.1.1.2

Multiply $- 1$ by $1$ .

$0 = - 1 + C - 2 C$

$B = 1 - C$

$A = - C$

Step 1.3.4.2.1.1.3

Multiply $- - C$ .

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Step 1.3.4.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

$0 = - 1 + 1 C - 2 C$

$B = 1 - C$

$A = - C$

Step 1.3.4.2.1.1.3.2

Multiply $C$ by $1$ .

$0 = - 1 + C - 2 C$

$B = 1 - C$

$A = - C$

$0 = - 1 + C - 2 C$

$B = 1 - C$

$A = - C$

$0 = - 1 + C - 2 C$

$B = 1 - C$

$A = - C$

Step 1.3.4.2.1.2

Subtract $2 C$ from $C$ .

$0 = - 1 - C$

$B = 1 - C$

$A = - C$

$0 = - 1 - C$

$B = 1 - C$

$A = - C$

$0 = - 1 - C$

$B = 1 - C$

$A = - C$

$0 = - 1 - C$

$B = 1 - C$

$A = - C$

Step 1.3.5

Solve for $C$ in $0 = - 1 - C$ .

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Step 1.3.5.1

Rewrite the equation as $- 1 - C = 0$ .

$- 1 - C = 0$

$B = 1 - C$

$A = - C$

Step 1.3.5.2

Add $1$ to both sides of the equation.

$- C = 1$

$B = 1 - C$

$A = - C$

Step 1.3.5.3

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

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Step 1.3.5.3.1

Divide each term in $- C = 1$ by $- 1$ .

$\frac{- C}{- 1} = \frac{1}{- 1}$

$B = 1 - C$

$A = - C$

Step 1.3.5.3.2

Simplify the left side.

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Step 1.3.5.3.2.1

Dividing two negative values results in a positive value.

$\frac{C}{1} = \frac{1}{- 1}$

$B = 1 - C$

$A = - C$

Step 1.3.5.3.2.2

Divide $C$ by $1$ .

$C = \frac{1}{- 1}$

$B = 1 - C$

$A = - C$

$C = \frac{1}{- 1}$

$B = 1 - C$

$A = - C$

Step 1.3.5.3.3

Simplify the right side.

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Step 1.3.5.3.3.1

Divide $1$ by $- 1$ .

$C = - 1$

$B = 1 - C$

$A = - C$

$C = - 1$

$B = 1 - C$

$A = - C$

$C = - 1$

$B = 1 - C$

$A = - C$

$C = - 1$

$B = 1 - C$

$A = - C$

Step 1.3.6

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

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Step 1.3.6.1

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

$B = 1 - (- 1)$

$C = - 1$

$A = - C$

Step 1.3.6.2

Simplify the right side.

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Step 1.3.6.2.1

Simplify $1 - (- 1)$ .

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Step 1.3.6.2.1.1

Multiply $- 1$ by $- 1$ .

$B = 1 + 1$

$C = - 1$

$A = - C$

Step 1.3.6.2.1.2

Add $1$ and $1$ .

$B = 2$

$C = - 1$

$A = - C$

$B = 2$

$C = - 1$

$A = - C$

$B = 2$

$C = - 1$

$A = - C$

Step 1.3.6.3

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

$A = - (- 1)$

$B = 2$

$C = - 1$

Step 1.3.6.4

Simplify the right side.

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Step 1.3.6.4.1

Multiply $- 1$ by $- 1$ .

$A = 1$

$B = 2$

$C = - 1$

$A = 1$

$B = 2$

$C = - 1$

$A = 1$

$B = 2$

$C = - 1$

Step 1.3.7

List all of the solutions.

$A = 1, B = 2, C = - 1$

Step 1.4

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

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

Step 1.5

Move the negative in front of the fraction.

$\int \frac{1}{z - 1} + \frac{2}{{(z - 2)}^{2}} - \frac{1}{z - 2} d z$

Step 2

Split the single integral into multiple integrals.

$\int \frac{1}{z - 1} d z + \int \frac{2}{{(z - 2)}^{2}} d z + \int - \frac{1}{z - 2} d z$

Step 3

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

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Step 3.1

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

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Step 3.1.1

Differentiate $z - 1$ .

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

Step 3.1.2

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

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

Step 3.1.3

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

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

Step 3.1.4

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

$1 + 0$

Step 3.1.5

Add $1$ and $0$ .

$1$

Step 3.2

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

$\int \frac{1}{u_{1}} d u_{1} + \int \frac{2}{{(z - 2)}^{2}} d z + \int - \frac{1}{z - 2} d z$

Step 4

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

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

Step 5

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

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

Step 6

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

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Step 6.1

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

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Step 6.1.1

Differentiate $z - 2$ .

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

Step 6.1.2

By the Sum Rule, the derivative of $z - 2$ with respect to $z$ is $\frac{d}{d z} [z] + \frac{d}{d z} [- 2]$ .

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

Step 6.1.3

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

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

Step 6.1.4

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

$1 + 0$

Step 6.1.5

Add $1$ and $0$ .

$1$

Step 6.2

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

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

Step 7

Apply basic rules of exponents.

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Step 7.1

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

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

Step 7.2

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

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Step 7.2.1

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

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

Step 7.2.2

Multiply $2$ by $- 1$ .

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

Step 8

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

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

Step 9

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

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

Step 10

Let $u_{3} = z - 2$ . Then $d u_{3} = d z$ . Rewrite using $u_{3}$ and $d$ $u_{3}$ .

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Step 10.1

Let $u_{3} = z - 2$ . Find $\frac{d u_{3}}{d z}$ .

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Step 10.1.1

Differentiate $z - 2$ .

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

Step 10.1.2

By the Sum Rule, the derivative of $z - 2$ with respect to $z$ is $\frac{d}{d z} [z] + \frac{d}{d z} [- 2]$ .

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

Step 10.1.3

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

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

Step 10.1.4

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

$1 + 0$

Step 10.1.5

Add $1$ and $0$ .

$1$

Step 10.2

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

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

Step 11

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

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

Step 12

Simplify.

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

Step 13

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

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

Step 14

Substitute back in for each integration substitution variable.

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Step 14.1

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

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

Step 14.2

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

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

Step 14.3

Replace all occurrences of $u_{3}$ with $z - 2$ .

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