Calculus Examples

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

Step 1

Write $\frac{3 x - 2}{{(x - 2)}^{2}}$ as a function.

$f (x) = \frac{3 x - 2}{{(x - 2)}^{2}}$

Step 2

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Step 3

Set up the integral to solve.

$F (x) = \int \frac{3 x - 2}{{(x - 2)}^{2}} d x$

Step 4

Write the fraction using partial fraction decomposition.

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Step 4.1

Decompose the fraction and multiply through by the common denominator.

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Step 4.1.1

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 4.1.2

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

Step 4.1.3

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

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

Step 4.1.4

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

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Step 4.1.4.1

Cancel the common factor.

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

Step 4.1.4.2

Divide $3 x - 2$ by $1$ .

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

Step 4.1.5

Simplify each term.

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Step 4.1.5.1

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

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Step 4.1.5.1.1

Cancel the common factor.

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

Step 4.1.5.1.2

Divide $A$ by $1$ .

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

Step 4.1.5.2

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

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Step 4.1.5.2.1

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

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

Step 4.1.5.2.2

Cancel the common factors.

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Step 4.1.5.2.2.1

Multiply by $1$ .

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

Step 4.1.5.2.2.2

Cancel the common factor.

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

Step 4.1.5.2.2.3

Rewrite the expression.

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

Step 4.1.5.2.2.4

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

$3 x - 2 = A + B (x - 2)$

Step 4.1.5.3

Apply the distributive property.

$3 x - 2 = A + B x + B \cdot - 2$

Step 4.1.5.4

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

$3 x - 2 = A + B x - 2 B$

Step 4.1.6

Reorder $A$ and $B x$ .

$3 x - 2 = B x + A - 2 B$

Step 4.2

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

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Step 4.2.1

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.

$3 = B$

Step 4.2.2

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.

$- 2 = A - 2 B$

Step 4.2.3

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

$3 = B$

$- 2 = A - 2 B$

$3 = B$

$- 2 = A - 2 B$

Step 4.3

Solve the system of equations.

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Step 4.3.1

Rewrite the equation as $B = 3$ .

$B = 3$

$- 2 = A - 2 B$

Step 4.3.2

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

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Step 4.3.2.1

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

$- 2 = A - 2 \cdot 3$

$B = 3$

Step 4.3.2.2

Simplify the right side.

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Step 4.3.2.2.1

Multiply $- 2$ by $3$ .

$- 2 = A - 6$

$B = 3$

$- 2 = A - 6$

$B = 3$

$- 2 = A - 6$

$B = 3$

Step 4.3.3

Solve for $A$ in $- 2 = A - 6$ .

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Step 4.3.3.1

Rewrite the equation as $A - 6 = - 2$ .

$A - 6 = - 2$

$B = 3$

Step 4.3.3.2

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

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Step 4.3.3.2.1

Add $6$ to both sides of the equation.

$A = - 2 + 6$

$B = 3$

Step 4.3.3.2.2

Add $- 2$ and $6$ .

$A = 4$

$B = 3$

$A = 4$

$B = 3$

$A = 4$

$B = 3$

Step 4.3.4

Solve the system of equations.

$A = 4 B = 3$

Step 4.3.5

List all of the solutions.

$A = 4, B = 3$

Step 4.4

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

$\frac{4}{{(x - 2)}^{2}} + \frac{3}{x - 2}$

Step 4.5

Remove the zero from the expression.

$\int \frac{4}{{(x - 2)}^{2}} + \frac{3}{x - 2} d x$

Step 5

Split the single integral into multiple integrals.

$\int \frac{4}{{(x - 2)}^{2}} d x + \int \frac{3}{x - 2} d x$

Step 6

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

$4 \int \frac{1}{{(x - 2)}^{2}} d x + \int \frac{3}{x - 2} d x$

Step 7

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

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Step 7.1

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

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Step 7.1.1

Differentiate $x - 2$ .

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

Step 7.1.2

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 7.1.3

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 7.1.4

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

$1 + 0$

Step 7.1.5

Add $1$ and $0$ .

$1$

Step 7.2

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

$4 \int \frac{1}{{u_{1}}^{2}} d u_{1} + \int \frac{3}{x - 2} d x$

Step 8

Apply basic rules of exponents.

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Step 8.1

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

$4 \int {({u_{1}}^{2})}^{- 1} d u_{1} + \int \frac{3}{x - 2} d x$

Step 8.2

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

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Step 8.2.1

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

$4 \int {u_{1}}^{2 \cdot - 1} d u_{1} + \int \frac{3}{x - 2} d x$

Step 8.2.2

Multiply $2$ by $- 1$ .

$4 \int {u_{1}}^{- 2} d u_{1} + \int \frac{3}{x - 2} d x$

Step 9

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

$4 (- {u_{1}}^{- 1} + C) + \int \frac{3}{x - 2} d x$

Step 10

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

$4 (- {u_{1}}^{- 1} + C) + 3 \int \frac{1}{x - 2} d x$

Step 11

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

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Step 11.1

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

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Step 11.1.1

Differentiate $x - 2$ .

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

Step 11.1.2

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 11.1.3

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 11.1.4

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

$1 + 0$

Step 11.1.5

Add $1$ and $0$ .

$1$

Step 11.2

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

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

Step 12

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

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

Step 13

Simplify.

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Step 13.1

Simplify.

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

Step 13.2

Simplify.

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Step 13.2.1

Multiply $- 1$ by $4$ .

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

Step 13.2.2

Combine $- 4$ and $\frac{1}{u_{1}}$ .

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

Step 13.2.3

Move the negative in front of the fraction.

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

Step 14

Substitute back in for each integration substitution variable.

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Step 14.1

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

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

Step 14.2

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

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

Step 15

The answer is the antiderivative of the function $f (x) = \frac{3 x - 2}{{(x - 2)}^{2}}$ .

$F (x) =$ $- \frac{4}{x - 2} + 3 \ln (| x - 2 |) + C$