Calculus Examples

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

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 the fraction.

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Step 1.1.1.1

Factor $x$ out of $x^{3} - 2 x^{2} + x$ .

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Step 1.1.1.1.1

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

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

Step 1.1.1.1.2

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

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

Step 1.1.1.1.3

Raise $x$ to the power of $1$ .

$\frac{2 x^{2} + 3}{x \cdot x^{2} + x (- 2 x) + x^{1}}$

Step 1.1.1.1.4

Factor $x$ out of $x^{1}$ .

$\frac{2 x^{2} + 3}{x \cdot x^{2} + x (- 2 x) + x \cdot 1}$

Step 1.1.1.1.5

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

$\frac{2 x^{2} + 3}{x (x^{2} - 2 x) + x \cdot 1}$

Step 1.1.1.1.6

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

$\frac{2 x^{2} + 3}{x (x^{2} - 2 x + 1)}$

Step 1.1.1.2

Factor using the perfect square rule.

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Step 1.1.1.2.1

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

$\frac{2 x^{2} + 3}{x (x^{2} - 2 x + 1^{2})}$

Step 1.1.1.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Step 1.1.1.2.3

Rewrite the polynomial.

$\frac{2 x^{2} + 3}{x (x^{2} - 2 \cdot x \cdot 1 + 1^{2})}$

Step 1.1.1.2.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

$\frac{2 x^{2} + 3}{x {(x - 1)}^{2}}$

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 $B$ .

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

Step 1.1.3

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

Step 1.1.4

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

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

Step 1.1.5

Cancel the common factor of $x$ .

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Step 1.1.5.1

Cancel the common factor.

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

Step 1.1.5.2

Rewrite the expression.

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

Step 1.1.6

Cancel the common factor of ${(x - 1)}^{2}$ .

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Step 1.1.6.1

Cancel the common factor.

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

Step 1.1.6.2

Divide $2 x^{2} + 3$ by $1$ .

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

Step 1.1.7

Simplify each term.

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Step 1.1.7.1

Cancel the common factor of $x$ .

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Step 1.1.7.1.1

Cancel the common factor.

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

Step 1.1.7.1.2

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

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

Step 1.1.7.2

Rewrite ${(x - 1)}^{2}$ as $(x - 1) (x - 1)$ .

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

Step 1.1.7.3

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

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Step 1.1.7.3.1

Apply the distributive property.

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

Step 1.1.7.3.2

Apply the distributive property.

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

Step 1.1.7.3.3

Apply the distributive property.

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

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 $x$ by $x$ .

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

Step 1.1.7.4.1.2

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

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

Step 1.1.7.4.1.3

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

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

Step 1.1.7.4.1.4

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

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

Step 1.1.7.4.1.5

Multiply $- 1$ by $- 1$ .

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

Step 1.1.7.4.2

Subtract $x$ from $- x$ .

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

Step 1.1.7.5

Apply the distributive property.

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

Step 1.1.7.6

Simplify.

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Step 1.1.7.6.1

Rewrite using the commutative property of multiplication.

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

Step 1.1.7.6.2

Multiply $A$ by $1$ .

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

Step 1.1.7.7

Cancel the common factor of ${(x - 1)}^{2}$ .

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Step 1.1.7.7.1

Cancel the common factor.

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

Step 1.1.7.7.2

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

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

Step 1.1.7.8

Cancel the common factor of ${(x - 1)}^{2}$ and $x - 1$ .

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Step 1.1.7.8.1

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

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

Step 1.1.7.8.2

Cancel the common factors.

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Step 1.1.7.8.2.1

Multiply by $1$ .

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

Step 1.1.7.8.2.2

Cancel the common factor.

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

Step 1.1.7.8.2.3

Rewrite the expression.

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

Step 1.1.7.8.2.4

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

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

Step 1.1.7.9

Apply the distributive property.

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

Step 1.1.7.10

Multiply $x$ by $x$ .

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

Step 1.1.7.11

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

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

Step 1.1.7.12

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

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

Step 1.1.7.13

Apply the distributive property.

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

Step 1.1.7.14

Rewrite using the commutative property of multiplication.

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

Step 1.1.8

Simplify the expression.

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Step 1.1.8.1

Move $A$ .

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

Step 1.1.8.2

Reorder $B$ and $x$ .

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

Step 1.1.8.3

Move $A$ .

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

Step 1.1.8.4

Move $B x$ .

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

Step 1.1.8.5

Move $- 2 x A$ .

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

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.

$2 = A + C$

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.

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

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.

$3 = A$

Step 1.2.4

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

$2 = A + C$

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

$3 = A$

$2 = A + C$

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

$3 = A$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Rewrite the equation as $A = 3$ .

$A = 3$

$2 = A + C$

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

Step 1.3.2

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

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Step 1.3.2.1

Replace all occurrences of $A$ in $2 = A + C$ with $3$ .

$2 = (3) + C$

$A = 3$

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

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Remove parentheses.

$2 = 3 + C$

$A = 3$

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

$2 = 3 + C$

$A = 3$

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

Step 1.3.2.3

Replace all occurrences of $A$ in $0 = - 2 A + B - 1 C$ with $3$ .

$0 = - 2 \cdot 3 + B - 1 C$

$2 = 3 + C$

$A = 3$

Step 1.3.2.4

Simplify the right side.

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Step 1.3.2.4.1

Simplify each term.

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Step 1.3.2.4.1.1

Multiply $- 2$ by $3$ .

$0 = - 6 + B - 1 C$

$2 = 3 + C$

$A = 3$

Step 1.3.2.4.1.2

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

$0 = - 6 + B - C$

$2 = 3 + C$

$A = 3$

$0 = - 6 + B - C$

$2 = 3 + C$

$A = 3$

$0 = - 6 + B - C$

$2 = 3 + C$

$A = 3$

$0 = - 6 + B - C$

$2 = 3 + C$

$A = 3$

Step 1.3.3

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

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Step 1.3.3.1

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

$3 + C = 2$

$0 = - 6 + B - C$

$A = 3$

Step 1.3.3.2

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

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Step 1.3.3.2.1

Subtract $3$ from both sides of the equation.

$C = 2 - 3$

$0 = - 6 + B - C$

$A = 3$

Step 1.3.3.2.2

Subtract $3$ from $2$ .

$C = - 1$

$0 = - 6 + B - C$

$A = 3$

$C = - 1$

$0 = - 6 + B - C$

$A = 3$

$C = - 1$

$0 = - 6 + B - C$

$A = 3$

Step 1.3.4

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

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Step 1.3.4.1

Replace all occurrences of $C$ in $0 = - 6 + B - C$ with $- 1$ .

$0 = - 6 + B - (- 1)$

$C = - 1$

$A = 3$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

Simplify $- 6 + B - (- 1)$ .

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Step 1.3.4.2.1.1

Multiply $- 1$ by $- 1$ .

$0 = - 6 + B + 1$

$C = - 1$

$A = 3$

Step 1.3.4.2.1.2

Add $- 6$ and $1$ .

$0 = B - 5$

$C = - 1$

$A = 3$

$0 = B - 5$

$C = - 1$

$A = 3$

$0 = B - 5$

$C = - 1$

$A = 3$

$0 = B - 5$

$C = - 1$

$A = 3$

Step 1.3.5

Solve for $B$ in $0 = B - 5$ .

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Step 1.3.5.1

Rewrite the equation as $B - 5 = 0$ .

$B - 5 = 0$

$C = - 1$

$A = 3$

Step 1.3.5.2

Add $5$ to both sides of the equation.

$B = 5$

$C = - 1$

$A = 3$

$B = 5$

$C = - 1$

$A = 3$

Step 1.3.6

Solve the system of equations.

$B = 5 C = - 1 A = 3$

Step 1.3.7

List all of the solutions.

$B = 5, C = - 1, A = 3$

Step 1.4

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

$\frac{3}{x} + \frac{5}{{(x - 1)}^{2}} + \frac{- 1}{x - 1}$

Step 1.5

Move the negative in front of the fraction.

$\int \frac{3}{x} + \frac{5}{{(x - 1)}^{2}} - \frac{1}{x - 1} d x$

Step 2

Split the single integral into multiple integrals.

$\int \frac{3}{x} d x + \int \frac{5}{{(x - 1)}^{2}} d x + \int - \frac{1}{x - 1} d x$

Step 3

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

$3 \int \frac{1}{x} d x + \int \frac{5}{{(x - 1)}^{2}} d x + \int - \frac{1}{x - 1} d x$

Step 4

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

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

Step 5

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

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

Step 6

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

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Step 6.1

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

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Step 6.1.1

Differentiate $x - 1$ .

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

Step 6.1.2

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 6.1.3

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

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

Step 6.1.4

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

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

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.

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

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, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 7.2.2

Multiply $2$ by $- 1$ .

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

Step 8

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

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

Step 9

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

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

Step 10

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

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Step 10.1

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

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Step 10.1.1

Differentiate $x - 1$ .

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

Step 10.1.2

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 10.1.3

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

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

Step 10.1.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ 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 $d u_{2}$ .

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

Step 11

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

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

Step 12

Simplify.

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

Step 13

Substitute back in for each integration substitution variable.

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Step 13.1

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

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

Step 13.2

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

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