Calculus Examples

$\frac{4}{x^{2} + 2 x + 1}$

Step 1

Write $\frac{4}{x^{2} + 2 x + 1}$ as a function.

$f (x) = \frac{4}{x^{2} + 2 x + 1}$

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.

$F (x) = \int \frac{4}{x^{2} + 2 x + 1} d x$

Step 4

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

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

Step 5

Write the fraction using partial fraction decomposition.

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Step 5.1

Decompose the fraction and multiply through by the common denominator.

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Step 5.1.1

Factor using the perfect square rule.

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Step 5.1.1.1

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

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

Step 5.1.1.2

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

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

Step 5.1.1.3

Rewrite the polynomial.

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

Step 5.1.1.4

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

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

Step 5.1.2

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 5.1.3

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

Step 5.1.4

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

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

Step 5.1.5

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

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Step 5.1.5.1

Cancel the common factor.

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

Step 5.1.5.2

Rewrite the expression.

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

Step 5.1.6

Simplify each term.

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Step 5.1.6.1

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

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Step 5.1.6.1.1

Cancel the common factor.

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

Step 5.1.6.1.2

Divide $A$ by $1$ .

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

Step 5.1.6.2

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

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Step 5.1.6.2.1

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

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

Step 5.1.6.2.2

Cancel the common factors.

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Step 5.1.6.2.2.1

Multiply by $1$ .

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

Step 5.1.6.2.2.2

Cancel the common factor.

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

Step 5.1.6.2.2.3

Rewrite the expression.

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

Step 5.1.6.2.2.4

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

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

Step 5.1.6.3

Apply the distributive property.

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

Step 5.1.6.4

Multiply $B$ by $1$ .

$1 = A + B x + B$

Step 5.1.7

Reorder $A$ and $B x$ .

$1 = B x + A + B$

Step 5.2

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

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Step 5.2.1

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.

$0 = B$

Step 5.2.2

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.

$1 = A + B$

Step 5.2.3

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

$0 = B$

$1 = A + B$

$0 = B$

$1 = A + B$

Step 5.3

Solve the system of equations.

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Step 5.3.1

Rewrite the equation as $B = 0$ .

$B = 0$

$1 = A + B$

Step 5.3.2

Replace all occurrences of $B$ with $0$ in each equation.

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Step 5.3.2.1

Replace all occurrences of $B$ in $1 = A + B$ with $0$ .

$1 = A + 0$

$B = 0$

Step 5.3.2.2

Simplify $1 = A + 0$ .

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Step 5.3.2.2.1

Simplify the left side.

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Step 5.3.2.2.1.1

Remove parentheses.

$1 = A + 0$

$B = 0$

$1 = A + 0$

$B = 0$

Step 5.3.2.2.2

Simplify the right side.

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Step 5.3.2.2.2.1

Add $A$ and $0$ .

$1 = A$

$B = 0$

$1 = A$

$B = 0$

$1 = A$

$B = 0$

$1 = A$

$B = 0$

Step 5.3.3

Rewrite the equation as $A = 1$ .

$A = 1$

$B = 0$

Step 5.3.4

Solve the system of equations.

$A = 1 B = 0$

Step 5.3.5

List all of the solutions.

$A = 1, B = 0$

Step 5.4

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

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

Step 5.5

Simplify.

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Step 5.5.1

Divide $0$ by $x + 1$ .

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

Step 5.5.2

Remove the zero from the expression.

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

Step 6

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

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Step 6.1

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

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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 $u$ and $d u$ .

$4 \int \frac{1}{u^{2}} d u$

Step 7

Apply basic rules of exponents.

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Step 7.1

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

$4 \int {(u^{2})}^{- 1} d u$

Step 7.2

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

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Step 7.2.1

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

$4 \int u^{2 \cdot - 1} d u$

Step 7.2.2

Multiply $2$ by $- 1$ .

$4 \int u^{- 2} d u$

Step 8

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

$4 (- u^{- 1} + C)$

Step 9

Simplify.

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Step 9.1

Rewrite $4 (- u^{- 1} + C)$ as $4 (- \frac{1}{u}) + C$ .

$4 (- \frac{1}{u}) + C$

Step 9.2

Simplify.

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Step 9.2.1

Multiply $- 1$ by $4$ .

$- 4 \frac{1}{u} + C$

Step 9.2.2

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

$\frac{- 4}{u} + C$

Step 9.2.3

Move the negative in front of the fraction.

$- \frac{4}{u} + C$

Step 10

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

$- \frac{4}{x + 1} + C$

Step 11

The answer is the antiderivative of the function $f (x) = \frac{4}{x^{2} + 2 x + 1}$ .

$F (x) =$ $- \frac{4}{x + 1} + C$