Calculus Examples

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

Step 1

Subtract $2 x$ from $x$ .

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

Step 2

Write the fraction using partial fraction decomposition.

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Step 2.1

Decompose the fraction and multiply through by the common denominator.

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Step 2.1.1

Factor $3$ out of $3 x - 3$ .

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Step 2.1.1.1

Factor $3$ out of $3 x$ .

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

Step 2.1.1.2

Factor $3$ out of $- 3$ .

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

Step 2.1.1.3

Factor $3$ out of $3 (x) + 3 (- 1)$ .

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

Step 2.1.2

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 $A$ .

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

Step 2.1.3

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

Step 2.1.4

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

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

Step 2.1.5

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

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Step 2.1.5.1

Cancel the common factor.

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

Step 2.1.5.2

Divide $3 (x - 1)$ by $1$ .

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

Step 2.1.6

Apply the distributive property.

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

Step 2.1.7

Multiply $3$ by $- 1$ .

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

Step 2.1.8

Simplify each term.

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Step 2.1.8.1

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

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Step 2.1.8.1.1

Cancel the common factor.

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

Step 2.1.8.1.2

Divide $A$ by $1$ .

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

Step 2.1.8.2

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

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Step 2.1.8.2.1

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

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

Step 2.1.8.2.2

Cancel the common factors.

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Step 2.1.8.2.2.1

Multiply by $1$ .

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

Step 2.1.8.2.2.2

Cancel the common factor.

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

Step 2.1.8.2.2.3

Rewrite the expression.

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

Step 2.1.8.2.2.4

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

$3 x - 3 = A + B (- x + 1)$

Step 2.1.8.3

Apply the distributive property.

$3 x - 3 = A + B (- x) + B \cdot 1$

Step 2.1.8.4

Rewrite using the commutative property of multiplication.

$3 x - 3 = A - B x + B \cdot 1$

Step 2.1.8.5

Multiply $B$ by $1$ .

$3 x - 3 = A - B x + B$

Step 2.1.9

Simplify the expression.

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Step 2.1.9.1

Move $B$ .

$3 x - 3 = A - x B + B$

Step 2.1.9.2

Reorder $A$ and $- x B$ .

$3 x - 3 = - x B + A + B$

Step 2.2

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

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Step 2.2.1

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.

$3 = - 1 B$

Step 2.2.2

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.

$- 3 = A + B$

Step 2.2.3

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

$3 = - 1 B$

$- 3 = A + B$

$3 = - 1 B$

$- 3 = A + B$

Step 2.3

Solve the system of equations.

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Step 2.3.1

Solve for $B$ in $3 = - 1 B$ .

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Step 2.3.1.1

Rewrite the equation as $- 1 B = 3$ .

$- 1 B = 3$

$- 3 = A + B$

Step 2.3.1.2

Divide each term in $- 1 B = 3$ by $- 1$ and simplify.

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Step 2.3.1.2.1

Divide each term in $- 1 B = 3$ by $- 1$ .

$\frac{- 1 B}{- 1} = \frac{3}{- 1}$

$- 3 = A + B$

Step 2.3.1.2.2

Simplify the left side.

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Step 2.3.1.2.2.1

Dividing two negative values results in a positive value.

$\frac{B}{1} = \frac{3}{- 1}$

$- 3 = A + B$

Step 2.3.1.2.2.2

Divide $B$ by $1$ .

$B = \frac{3}{- 1}$

$- 3 = A + B$

$B = \frac{3}{- 1}$

$- 3 = A + B$

Step 2.3.1.2.3

Simplify the right side.

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Step 2.3.1.2.3.1

Divide $3$ by $- 1$ .

$B = - 3$

$- 3 = A + B$

$B = - 3$

$- 3 = A + B$

$B = - 3$

$- 3 = A + B$

$B = - 3$

$- 3 = A + B$

Step 2.3.2

Replace all occurrences of $B$ with $- 3$ in each equation.

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Step 2.3.2.1

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

$- 3 = A - 3$

$B = - 3$

Step 2.3.2.2

Simplify the left side.

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Step 2.3.2.2.1

Remove parentheses.

$- 3 = A - 3$

$B = - 3$

$- 3 = A - 3$

$B = - 3$

$- 3 = A - 3$

$B = - 3$

Step 2.3.3

Solve for $A$ in $- 3 = A - 3$ .

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Step 2.3.3.1

Rewrite the equation as $A - 3 = - 3$ .

$A - 3 = - 3$

$B = - 3$

Step 2.3.3.2

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

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Step 2.3.3.2.1

Add $3$ to both sides of the equation.

$A = - 3 + 3$

$B = - 3$

Step 2.3.3.2.2

Add $- 3$ and $3$ .

$A = 0$

$B = - 3$

$A = 0$

$B = - 3$

$A = 0$

$B = - 3$

Step 2.3.4

Solve the system of equations.

$A = 0 B = - 3$

Step 2.3.5

List all of the solutions.

$A = 0, B = - 3$

Step 2.4

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

$\frac{0}{{(- x + 1)}^{2}} + \frac{- 3}{- x + 1}$

Step 2.5

Simplify.

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Step 2.5.1

Divide $0$ by ${(- x + 1)}^{2}$ .

$0 + \frac{- 3}{- x + 1}$

Step 2.5.2

Move the negative in front of the fraction.

$0 - \frac{3}{- x + 1}$

Step 2.5.3

Factor $- 1$ out of $- x$ .

$0 - \frac{3}{- (x) + 1}$

Step 2.5.4

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

$0 - \frac{3}{- (x) - 1 \cdot - 1}$

Step 2.5.5

Factor $- 1$ out of $- (x) - 1 (- 1)$ .

$0 - \frac{3}{- (x - 1)}$

Step 2.5.6

Simplify the expression.

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Step 2.5.6.1

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

$0 - \frac{3}{- 1 (x - 1)}$

Step 2.5.6.2

Move the negative in front of the fraction.

$0 + \frac{3}{x - 1}$

Step 2.5.6.3

Multiply $- 1$ by $- 1$ .

$0 + 1 (\frac{3}{x - 1})$

Step 2.5.6.4

Multiply $\frac{3}{x - 1}$ by $1$ .

$0 + \frac{3}{x - 1}$

Step 2.5.7

Remove the zero from the expression.

$\int \frac{3}{x - 1} d x$

Step 3

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

$3 \int \frac{1}{x - 1} d x$

Step 4

Let $u = x - 1$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.1

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

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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 $u$ and $d u$ .

$3 \int \frac{1}{u} d u$

Step 5

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

$3 (\ln (| u |) + C)$

Step 6

Simplify.

$3 \ln (| u |) + C$

Step 7

Replace all occurrences of $u$ with $x - 1$ .

$3 \ln (| x - 1 |) + C$