Calculus Examples

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

Step 1

Write $\frac{2 x + 1}{{(x + 2)}^{2}}$ as a function.

$f (x) = \frac{2 x + 1}{{(x + 2)}^{2}}$

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.

$F (x) = \int \frac{2 x + 1}{{(x + 2)}^{2}} d x$

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 ${(x + 2)}^{2}$ .

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

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.

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

Step 4.1.4.2

Divide $2 x + 1$ by $1$ .

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

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.

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

Step 4.1.5.1.2

Divide $A$ by $1$ .

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

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 $(B) {(x + 2)}^{2}$ .

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

Step 4.1.5.2.2

Cancel the common factors.

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Step 4.1.5.2.2.1

Multiply by $1$ .

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

Step 4.1.5.2.2.2

Cancel the common factor.

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

Step 4.1.5.2.2.3

Rewrite the expression.

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

Step 4.1.5.2.2.4

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

$2 x + 1 = A + B (x + 2)$

Step 4.1.5.3

Apply the distributive property.

$2 x + 1 = A + B x + B \cdot 2$

Step 4.1.5.4

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

$2 x + 1 = A + B x + 2 B$

Step 4.1.6

Reorder $A$ and $B x$ .

$2 x + 1 = B x + A + 2 B$

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.

$2 = B$

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.

$1 = A + 2 B$

Step 4.2.3

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

$2 = B$

$1 = A + 2 B$

$2 = B$

$1 = A + 2 B$

Step 4.3

Solve the system of equations.

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Step 4.3.1

Rewrite the equation as $B = 2$ .

$B = 2$

$1 = A + 2 B$

Step 4.3.2

Replace all occurrences of $B$ with $2$ in each equation.

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Step 4.3.2.1

Replace all occurrences of $B$ in $1 = A + 2 B$ with $2$ .

$1 = A + 2 (2)$

$B = 2$

Step 4.3.2.2

Simplify the right side.

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Step 4.3.2.2.1

Multiply $2$ by $2$ .

$1 = A + 4$

$B = 2$

$1 = A + 4$

$B = 2$

$1 = A + 4$

$B = 2$

Step 4.3.3

Solve for $A$ in $1 = A + 4$ .

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Step 4.3.3.1

Rewrite the equation as $A + 4 = 1$ .

$A + 4 = 1$

$B = 2$

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

Subtract $4$ from both sides of the equation.

$A = 1 - 4$

$B = 2$

Step 4.3.3.2.2

Subtract $4$ from $1$ .

$A = - 3$

$B = 2$

$A = - 3$

$B = 2$

$A = - 3$

$B = 2$

Step 4.3.4

Solve the system of equations.

$A = - 3 B = 2$

Step 4.3.5

List all of the solutions.

$A = - 3, B = 2$

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 $B$ .

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

Step 4.5

Move the negative in front of the fraction.

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

Step 5

Split the single integral into multiple integrals.

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

Step 6

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

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

Step 7

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

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

Step 8

Multiply $3$ by $- 1$ .

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

Step 9

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

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Step 9.1

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

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Step 9.1.1

Differentiate $x + 2$ .

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

Step 9.1.2

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 9.1.3

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

$1 + \frac{d}{d x} [2]$

Step 9.1.4

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

$1 + 0$

Step 9.1.5

Add $1$ and $0$ .

$1$

Step 9.2

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

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

Step 10

Apply basic rules of exponents.

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Step 10.1

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

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

Step 10.2

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

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Step 10.2.1

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

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

Step 10.2.2

Multiply $2$ by $- 1$ .

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

Step 11

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

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

Step 12

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

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

Step 13

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

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Step 13.1

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

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Step 13.1.1

Differentiate $x + 2$ .

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

Step 13.1.2

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 13.1.3

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 13.1.4

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

$1 + 0$

Step 13.1.5

Add $1$ and $0$ .

$1$

Step 13.2

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

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

Step 14

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

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

Step 15

Simplify.

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Step 15.1

Simplify.

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

Step 15.2

Simplify.

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Step 15.2.1

Multiply $- 1$ by $- 3$ .

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

Step 15.2.2

Combine $3$ and $\frac{1}{u_{1}}$ .

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

Step 16

Substitute back in for each integration substitution variable.

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Step 16.1

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

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

Step 16.2

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

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

Step 17

The answer is the antiderivative of the function $f (x) = \frac{2 x + 1}{{(x + 2)}^{2}}$ .

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