Calculus Examples

$\frac{x - 1}{x^{2} + 6 x + 9}$

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 using the perfect square rule.

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Step 1.1.1.1

Rewrite $9$ as $3^{2}$ .

$\frac{x - 1}{x^{2} + 6 x + 3^{2}}$

Step 1.1.1.2

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

$6 x = 2 \cdot x \cdot 3$

Step 1.1.1.3

Rewrite the polynomial.

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

Step 1.1.1.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 3$ .

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

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 $A$ .

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

Step 1.1.3

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

Step 1.1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is ${(x + 3)}^{2}$ .

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

Step 1.1.5

Cancel the common factor of ${(x + 3)}^{2}$ .

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Step 1.1.5.1

Cancel the common factor.

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

Step 1.1.5.2

Divide $x - 1$ by $1$ .

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

Step 1.1.6

Simplify each term.

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Step 1.1.6.1

Cancel the common factor of ${(x + 3)}^{2}$ .

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Step 1.1.6.1.1

Cancel the common factor.

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

Step 1.1.6.1.2

Divide $A$ by $1$ .

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

Step 1.1.6.2

Cancel the common factor of ${(x + 3)}^{2}$ and $x + 3$ .

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Step 1.1.6.2.1

Factor $x + 3$ out of $(B) {(x + 3)}^{2}$ .

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

Step 1.1.6.2.2

Cancel the common factors.

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Step 1.1.6.2.2.1

Multiply by $1$ .

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

Step 1.1.6.2.2.2

Cancel the common factor.

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

Step 1.1.6.2.2.3

Rewrite the expression.

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

Step 1.1.6.2.2.4

Divide $B (x + 3)$ by $1$ .

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

Step 1.1.6.3

Apply the distributive property.

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

Step 1.1.6.4

Move $3$ to the left of $B$ .

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

Step 1.1.7

Reorder $A$ and $B x$ .

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

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

$1 = B$

Step 1.2.2

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.

$- 1 = A + 3 B$

Step 1.2.3

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

$1 = B$

$- 1 = A + 3 B$

$1 = B$

$- 1 = A + 3 B$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Rewrite the equation as $B = 1$ .

$B = 1$

$- 1 = A + 3 B$

Step 1.3.2

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

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Step 1.3.2.1

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

$- 1 = A + 3 (1)$

$B = 1$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Multiply $3$ by $1$ .

$- 1 = A + 3$

$B = 1$

$- 1 = A + 3$

$B = 1$

$- 1 = A + 3$

$B = 1$

Step 1.3.3

Solve for $A$ in $- 1 = A + 3$ .

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Step 1.3.3.1

Rewrite the equation as $A + 3 = - 1$ .

$A + 3 = - 1$

$B = 1$

Step 1.3.3.2

Move all terms not containing $A$ 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.

$A = - 1 - 3$

$B = 1$

Step 1.3.3.2.2

Subtract $3$ from $- 1$ .

$A = - 4$

$B = 1$

$A = - 4$

$B = 1$

$A = - 4$

$B = 1$

Step 1.3.4

Solve the system of equations.

$A = - 4 B = 1$

Step 1.3.5

List all of the solutions.

$A = - 4, B = 1$

Step 1.4

Replace each of the partial fraction coefficients in $\frac{A}{{(x + 3)}^{2}} + \frac{B}{x + 3}$ with the values found for $A$ and $B$ .

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

Step 1.5

Move the negative in front of the fraction.

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

Step 2

Split the single integral into multiple integrals.

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

Step 3

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

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

Step 4

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

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

Step 5

Multiply $4$ by $- 1$ .

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

Step 6

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

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Step 6.1

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

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Step 6.1.1

Differentiate $x + 3$ .

$\frac{d}{d x} [x + 3]$

Step 6.1.2

By the Sum Rule, the derivative of $x + 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [3]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [3]$

Step 6.1.3

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

$1 + \frac{d}{d x} [3]$

Step 6.1.4

Since $3$ is constant with respect to $x$ , the derivative of $3$ 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 $d u_{1}$ .

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

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.

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

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

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

Step 7.2.2

Multiply $2$ by $- 1$ .

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

Step 8

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

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

Step 9

Let $u_{2} = x + 3$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 9.1

Let $u_{2} = x + 3$ . Find $\frac{d u_{2}}{d x}$ .

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Step 9.1.1

Differentiate $x + 3$ .

$\frac{d}{d x} [x + 3]$

Step 9.1.2

By the Sum Rule, the derivative of $x + 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [3]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [3]$

Step 9.1.3

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

$1 + \frac{d}{d x} [3]$

Step 9.1.4

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$1 + 0$

Step 9.1.5

Add $1$ and $0$ .

$1$

Step 9.2

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

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

Step 10

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

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

Step 11

Simplify.

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Step 11.1

Simplify.

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

Step 11.2

Simplify.

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Step 11.2.1

Multiply $- 1$ by $- 4$ .

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

Step 11.2.2

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

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

Step 12

Substitute back in for each integration substitution variable.

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Step 12.1

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

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

Step 12.2

Replace all occurrences of $u_{2}$ with $x + 3$ .

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