Calculus Examples

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

Step 1

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

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

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.

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

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 $C$ .

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

Step 4.1.2

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

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

Step 4.1.3

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

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Step 4.1.3.1

Cancel the common factor.

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

Step 4.1.3.2

Rewrite the expression.

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

Step 4.1.4

Cancel the common factor of $x + 1$ .

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Step 4.1.4.1

Cancel the common factor.

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

Step 4.1.4.2

Rewrite the expression.

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

Step 4.1.5

Simplify each term.

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Step 4.1.5.1

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

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Step 4.1.5.1.1

Cancel the common factor.

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

Step 4.1.5.1.2

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

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

Step 4.1.5.2

Apply the distributive property.

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

Step 4.1.5.3

Multiply $A$ by $1$ .

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

Step 4.1.5.4

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

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Step 4.1.5.4.1

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

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

Step 4.1.5.4.2

Cancel the common factors.

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Step 4.1.5.4.2.1

Raise $x$ to the power of $1$ .

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

Step 4.1.5.4.2.2

Factor $x$ out of $x^{1}$ .

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

Step 4.1.5.4.2.3

Cancel the common factor.

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

Step 4.1.5.4.2.4

Rewrite the expression.

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

Step 4.1.5.4.2.5

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

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

Step 4.1.5.5

Apply the distributive property.

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

Step 4.1.5.6

Multiply $x$ by $x$ .

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

Step 4.1.5.7

Multiply $x$ by $1$ .

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

Step 4.1.5.8

Apply the distributive property.

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

Step 4.1.5.9

Cancel the common factor of $x + 1$ .

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Step 4.1.5.9.1

Cancel the common factor.

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

Step 4.1.5.9.2

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

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

$1 = A x + A + B x^{2} + B x + C x^{2}$

Step 4.1.6

Simplify the expression.

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Step 4.1.6.1

Reorder $B$ and $x$ .

$1 = A x + A + B x^{2} + B x + C x^{2}$

Step 4.1.6.2

Move $A$ .

$1 = A x + B x^{2} + B x + C x^{2} + A$

Step 4.1.6.3

Move $B x$ .

$1 = A x + B x^{2} + C x^{2} + B x + A$

Step 4.1.6.4

Move $A x$ .

$1 = B x^{2} + C x^{2} + A x + B x + A$

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 $x^{2}$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = B + C$

Step 4.2.2

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.

$0 = A + B$

Step 4.2.3

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.

$1 = A$

Step 4.2.4

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

$0 = B + C$

$0 = A + B$

$1 = A$

$0 = B + C$

$0 = A + B$

$1 = A$

Step 4.3

Solve the system of equations.

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Step 4.3.1

Rewrite the equation as $A = 1$ .

$A = 1$

$0 = B + C$

$0 = A + B$

Step 4.3.2

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

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Step 4.3.2.1

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

$0 = (1) + B$

$A = 1$

$0 = B + C$

Step 4.3.2.2

Simplify the right side.

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Step 4.3.2.2.1

Remove parentheses.

$0 = 1 + B$

$A = 1$

$0 = B + C$

$0 = 1 + B$

$A = 1$

$0 = B + C$

$0 = 1 + B$

$A = 1$

$0 = B + C$

Step 4.3.3

Solve for $B$ in $0 = 1 + B$ .

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Step 4.3.3.1

Rewrite the equation as $1 + B = 0$ .

$1 + B = 0$

$A = 1$

$0 = B + C$

Step 4.3.3.2

Subtract $1$ from both sides of the equation.

$B = - 1$

$A = 1$

$0 = B + C$

$B = - 1$

$A = 1$

$0 = B + C$

Step 4.3.4

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

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Step 4.3.4.1

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

$0 = (- 1) + C$

$B = - 1$

$A = 1$

Step 4.3.4.2

Simplify the right side.

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Step 4.3.4.2.1

Remove parentheses.

$0 = - 1 + C$

$B = - 1$

$A = 1$

$0 = - 1 + C$

$B = - 1$

$A = 1$

$0 = - 1 + C$

$B = - 1$

$A = 1$

Step 4.3.5

Solve for $C$ in $0 = - 1 + C$ .

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Step 4.3.5.1

Rewrite the equation as $- 1 + C = 0$ .

$- 1 + C = 0$

$B = - 1$

$A = 1$

Step 4.3.5.2

Add $1$ to both sides of the equation.

$C = 1$

$B = - 1$

$A = 1$

$C = 1$

$B = - 1$

$A = 1$

Step 4.3.6

Solve the system of equations.

$C = 1 B = - 1 A = 1$

Step 4.3.7

List all of the solutions.

$C = 1, B = - 1, A = 1$

Step 4.4

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

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

Step 4.5

Move the negative in front of the fraction.

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

Step 5

Split the single integral into multiple integrals.

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

Step 6

Apply basic rules of exponents.

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Step 6.1

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

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

Step 6.2

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

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Step 6.2.1

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

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

Step 6.2.2

Multiply $2$ by $- 1$ .

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

Step 7

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

$- x^{- 1} + C + \int - \frac{1}{x} d x + \int \frac{1}{x + 1} d x$

Step 8

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

$- x^{- 1} + C - \int \frac{1}{x} d x + \int \frac{1}{x + 1} d x$

Step 9

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

$- x^{- 1} + C - (\ln (| x |) + C) + \int \frac{1}{x + 1} d x$

Step 10

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

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Step 10.1

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

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Step 10.1.1

Differentiate $x + 1$ .

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

Step 10.1.2

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 10.1.3

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 10.1.4

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

$1 + 0$

Step 10.1.5

Add $1$ and $0$ .

$1$

Step 10.2

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

$- x^{- 1} + C - (\ln (| x |) + C) + \int \frac{1}{u} d u$

Step 11

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

$- x^{- 1} + C - (\ln (| x |) + C) + \ln (| u |) + C$

Step 12

Simplify.

$- \frac{1}{x} - \ln (| x |) + \ln (| u |) + C$

Step 13

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

$- \frac{1}{x} - \ln (| x |) + \ln (| x + 1 |) + C$

Step 14

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

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