Calculus Examples

$- \frac{48}{{(2 x + 2)}^{2}} + 48$

Step 1

Write $- \frac{48}{{(2 x + 2)}^{2}} + 48$ as a function.

$f (x) = - \frac{48}{{(2 x + 2)}^{2}} + 48$

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.

$F (x) = \int - \frac{48}{{(2 x + 2)}^{2}} + 48 d x$

Step 4

Split the single integral into multiple integrals.

$\int - \frac{48}{{(2 x + 2)}^{2}} d x + \int 48 d x$

Step 5

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

$- \int \frac{48}{{(2 x + 2)}^{2}} d x + \int 48 d x$

Step 6

Since $48$ is constant with respect to $x$ , move $48$ out of the integral.

$- (48 \int \frac{1}{{(2 x + 2)}^{2}} d x) + \int 48 d x$

Step 7

Multiply $48$ by $- 1$ .

$- 48 \int \frac{1}{{(2 x + 2)}^{2}} d x + \int 48 d x$

Step 8

Write the fraction using partial fraction decomposition.

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Step 8.1

Decompose the fraction and multiply through by the common denominator.

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Step 8.1.1

Factor the fraction.

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Step 8.1.1.1

Factor $2$ out of $2 x + 2$ .

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Step 8.1.1.1.1

Factor $2$ out of $2 x$ .

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

Step 8.1.1.1.2

Factor $2$ out of $2$ .

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

Step 8.1.1.1.3

Factor $2$ out of $2 (x) + 2 (1)$ .

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

Step 8.1.1.2

Apply the product rule to $2 (x + 1)$ .

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

Step 8.1.2

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 8.1.3

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

Step 8.1.4

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

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

Step 8.1.5

Cancel the common factor of $2^{2}$ .

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Step 8.1.5.1

Cancel the common factor.

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

Step 8.1.5.2

Rewrite the expression.

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

Step 8.1.6

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

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Step 8.1.6.1

Cancel the common factor.

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

Step 8.1.6.2

Rewrite the expression.

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

Step 8.1.7

Simplify each term.

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Step 8.1.7.1

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

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Step 8.1.7.1.1

Cancel the common factor.

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

Step 8.1.7.1.2

Divide $(A) \cdot (2^{2})$ by $1$ .

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

Step 8.1.7.2

Raise $2$ to the power of $2$ .

$1 = A \cdot 4 + \frac{(B) (2^{2} {(x + 1)}^{2})}{x + 1}$

Step 8.1.7.3

Move $4$ to the left of $A$ .

$1 = 4 \cdot A + \frac{(B) (2^{2} {(x + 1)}^{2})}{x + 1}$

Step 8.1.7.4

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

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Step 8.1.7.4.1

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

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

Step 8.1.7.4.2

Cancel the common factors.

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Step 8.1.7.4.2.1

Multiply by $1$ .

$1 = 4 A + \frac{(x + 1) (B (2^{2} (x + 1)))}{(x + 1) \cdot 1}$

Step 8.1.7.4.2.2

Cancel the common factor.

$1 = 4 A + \frac{(x + 1) (B (2^{2} (x + 1)))}{(x + 1) \cdot 1}$

Step 8.1.7.4.2.3

Rewrite the expression.

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

Step 8.1.7.4.2.4

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

$1 = 4 A + B (2^{2} (x + 1))$

Step 8.1.7.5

Rewrite using the commutative property of multiplication.

$1 = 4 A + 2^{2} B (x + 1)$

Step 8.1.7.6

Raise $2$ to the power of $2$ .

$1 = 4 A + 4 B (x + 1)$

Step 8.1.7.7

Apply the distributive property.

$1 = 4 A + 4 B x + 4 B \cdot 1$

Step 8.1.7.8

Multiply $4$ by $1$ .

$1 = 4 A + 4 B x + 4 B$

Step 8.1.8

Simplify the expression.

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Step 8.1.8.1

Move $B$ .

$1 = 4 A + 4 x B + 4 B$

Step 8.1.8.2

Reorder $4 A$ and $4 x B$ .

$1 = 4 x B + 4 A + 4 B$

Step 8.2

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

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Step 8.2.1

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.

$0 = 4 B$

Step 8.2.2

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.

$1 = 4 A + 4 B$

Step 8.2.3

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

$0 = 4 B$

$1 = 4 A + 4 B$

$0 = 4 B$

$1 = 4 A + 4 B$

Step 8.3

Solve the system of equations.

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Step 8.3.1

Solve for $B$ in $0 = 4 B$ .

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Step 8.3.1.1

Rewrite the equation as $4 B = 0$ .

$4 B = 0$

$1 = 4 A + 4 B$

Step 8.3.1.2

Divide each term in $4 B = 0$ by $4$ and simplify.

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Step 8.3.1.2.1

Divide each term in $4 B = 0$ by $4$ .

$\frac{4 B}{4} = \frac{0}{4}$

$1 = 4 A + 4 B$

Step 8.3.1.2.2

Simplify the left side.

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Step 8.3.1.2.2.1

Cancel the common factor of $4$ .

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Step 8.3.1.2.2.1.1

Cancel the common factor.

$\frac{4 B}{4} = \frac{0}{4}$

$1 = 4 A + 4 B$

Step 8.3.1.2.2.1.2

Divide $B$ by $1$ .

$B = \frac{0}{4}$

$1 = 4 A + 4 B$

$B = \frac{0}{4}$

$1 = 4 A + 4 B$

$B = \frac{0}{4}$

$1 = 4 A + 4 B$

Step 8.3.1.2.3

Simplify the right side.

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Step 8.3.1.2.3.1

Divide $0$ by $4$ .

$B = 0$

$1 = 4 A + 4 B$

$B = 0$

$1 = 4 A + 4 B$

$B = 0$

$1 = 4 A + 4 B$

$B = 0$

$1 = 4 A + 4 B$

Step 8.3.2

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

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Step 8.3.2.1

Replace all occurrences of $B$ in $1 = 4 A + 4 B$ with $0$ .

$1 = 4 A + 4 (0)$

$B = 0$

Step 8.3.2.2

Simplify the right side.

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Step 8.3.2.2.1

Simplify $4 A + 4 (0)$ .

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Step 8.3.2.2.1.1

Multiply $4$ by $0$ .

$1 = 4 A + 0$

$B = 0$

Step 8.3.2.2.1.2

Add $4 A$ and $0$ .

$1 = 4 A$

$B = 0$

$1 = 4 A$

$B = 0$

$1 = 4 A$

$B = 0$

$1 = 4 A$

$B = 0$

Step 8.3.3

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

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Step 8.3.3.1

Rewrite the equation as $4 A = 1$ .

$4 A = 1$

$B = 0$

Step 8.3.3.2

Divide each term in $4 A = 1$ by $4$ and simplify.

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Step 8.3.3.2.1

Divide each term in $4 A = 1$ by $4$ .

$\frac{4 A}{4} = \frac{1}{4}$

$B = 0$

Step 8.3.3.2.2

Simplify the left side.

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Step 8.3.3.2.2.1

Cancel the common factor of $4$ .

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Step 8.3.3.2.2.1.1

Cancel the common factor.

$\frac{4 A}{4} = \frac{1}{4}$

$B = 0$

Step 8.3.3.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{1}{4}$

$B = 0$

$A = \frac{1}{4}$

$B = 0$

$A = \frac{1}{4}$

$B = 0$

$A = \frac{1}{4}$

$B = 0$

$A = \frac{1}{4}$

$B = 0$

Step 8.3.4

Solve the system of equations.

$A = \frac{1}{4} B = 0$

Step 8.3.5

List all of the solutions.

$A = \frac{1}{4}, B = 0$

Step 8.4

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

$\frac{\frac{1}{4}}{{(x + 1)}^{2}} + \frac{0}{x + 1}$

Step 8.5

Simplify.

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Step 8.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{4} \cdot \frac{1}{{(x + 1)}^{2}} + \frac{0}{x + 1}$

Step 8.5.2

Combine.

$\frac{1 \cdot 1}{4 {(x + 1)}^{2}} + \frac{0}{x + 1}$

Step 8.5.3

Multiply $1$ by $1$ .

$\frac{1}{4 {(x + 1)}^{2}} + \frac{0}{x + 1}$

Step 8.5.4

Divide $0$ by $x + 1$ .

$\frac{1}{4 {(x + 1)}^{2}} + 0$

Step 8.5.5

Remove the zero from the expression.

$- 48 \int \frac{1}{4 {(x + 1)}^{2}} d x + \int 48 d x$

Step 9

Since $\frac{1}{4}$ is constant with respect to $x$ , move $\frac{1}{4}$ out of the integral.

$- 48 (\frac{1}{4} \int \frac{1}{{(x + 1)}^{2}} d x) + \int 48 d x$

Step 10

Simplify.

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Step 10.1

Combine $\frac{1}{4}$ and $- 48$ .

$\frac{- 48}{4} \int \frac{1}{{(x + 1)}^{2}} d x + \int 48 d x$

Step 10.2

Cancel the common factor of $- 48$ and $4$ .

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Step 10.2.1

Factor $4$ out of $- 48$ .

$\frac{4 \cdot - 12}{4} \int \frac{1}{{(x + 1)}^{2}} d x + \int 48 d x$

Step 10.2.2

Cancel the common factors.

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Step 10.2.2.1

Factor $4$ out of $4$ .

$\frac{4 \cdot - 12}{4 (1)} \int \frac{1}{{(x + 1)}^{2}} d x + \int 48 d x$

Step 10.2.2.2

Cancel the common factor.

$\frac{4 \cdot - 12}{4 \cdot 1} \int \frac{1}{{(x + 1)}^{2}} d x + \int 48 d x$

Step 10.2.2.3

Rewrite the expression.

$\frac{- 12}{1} \int \frac{1}{{(x + 1)}^{2}} d x + \int 48 d x$

Step 10.2.2.4

Divide $- 12$ by $1$ .

$- 12 \int \frac{1}{{(x + 1)}^{2}} d x + \int 48 d x$

Step 11

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

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Step 11.1

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

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Step 11.1.1

Differentiate $x + 1$ .

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

Step 11.1.2

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 11.1.3

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

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

Step 11.1.4

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$1 + 0$

Step 11.1.5

Add $1$ and $0$ .

$1$

Step 11.2

Rewrite the problem using $u$ and $d u$ .

$- 12 \int \frac{1}{u^{2}} d u + \int 48 d x$

Step 12

Apply basic rules of exponents.

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Step 12.1

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

$- 12 \int {(u^{2})}^{- 1} d u + \int 48 d x$

Step 12.2

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

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Step 12.2.1

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

$- 12 \int u^{2 \cdot - 1} d u + \int 48 d x$

Step 12.2.2

Multiply $2$ by $- 1$ .

$- 12 \int u^{- 2} d u + \int 48 d x$

Step 13

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

$- 12 (- u^{- 1} + C) + \int 48 d x$

Step 14

Apply the constant rule.

$- 12 (- u^{- 1} + C) + 48 x + C$

Step 15

Simplify.

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Step 15.1

Simplify.

$- 12 (- \frac{1}{u}) + 48 x + C$

Step 15.2

Simplify.

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Step 15.2.1

Multiply $- 1$ by $- 12$ .

$12 \frac{1}{u} + 48 x + C$

Step 15.2.2

Combine $12$ and $\frac{1}{u}$ .

$\frac{12}{u} + 48 x + C$

Step 16

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

$\frac{12}{x + 1} + 48 x + C$

Step 17

The answer is the antiderivative of the function $f (x) = - \frac{48}{{(2 x + 2)}^{2}} + 48$ .

$F (x) =$ $\frac{12}{x + 1} + 48 x + C$