Calculus Examples

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

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 the fraction.

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Step 1.1.1.1

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

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

Step 1.1.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

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

Step 1.1.1.3

Factor $x$ out of $x^{3} + x$ .

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Step 1.1.1.3.1

Factor $x$ out of $x^{3}$ .

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

Step 1.1.1.3.2

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

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

Step 1.1.1.3.3

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

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

Step 1.1.1.3.4

Factor $x$ out of $x \cdot x^{2} + x \cdot 1$ .

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

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 is 2nd order, $2$ terms are required in the numerator. The number of terms required in the numerator is always equal to the order of the factor in the denominator.

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

Step 1.1.3

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

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

Step 1.1.4

Reduce the expression by cancelling the common factors.

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Step 1.1.4.1

Cancel the common factor of $x$ .

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Step 1.1.4.1.1

Cancel the common factor.

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

Step 1.1.4.1.2

Rewrite the expression.

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

Step 1.1.4.2

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

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Step 1.1.4.2.1

Cancel the common factor.

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

Step 1.1.4.2.2

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

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

Step 1.1.5

Expand $(x + 1) (x - 1)$ using the FOIL Method.

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Step 1.1.5.1

Apply the distributive property.

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

Step 1.1.5.2

Apply the distributive property.

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

Step 1.1.5.3

Apply the distributive property.

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

Step 1.1.6

Simplify and combine like terms.

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Step 1.1.6.1

Simplify each term.

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Step 1.1.6.1.1

Multiply $x$ by $x$ .

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

Step 1.1.6.1.2

Move $- 1$ to the left of $x$ .

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

Step 1.1.6.1.3

Rewrite $- 1 x$ as $- x$ .

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

Step 1.1.6.1.4

Multiply $x$ by $1$ .

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

Step 1.1.6.1.5

Multiply $- 1$ by $1$ .

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

Step 1.1.6.2

Add $- x$ and $x$ .

$x^{2} + 0 - 1 = \frac{A (x (x^{2} + 1))}{x} + \frac{(B x + C) (x (x^{2} + 1))}{x^{2} + 1}$

Step 1.1.6.3

Add $x^{2}$ and $0$ .

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

Step 1.1.7

Simplify each term.

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Step 1.1.7.1

Cancel the common factor of $x$ .

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Step 1.1.7.1.1

Cancel the common factor.

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

Step 1.1.7.1.2

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

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

Step 1.1.7.2

Apply the distributive property.

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

Step 1.1.7.3

Multiply $A$ by $1$ .

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

Step 1.1.7.4

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

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Step 1.1.7.4.1

Cancel the common factor.

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

Step 1.1.7.4.2

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

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

Step 1.1.7.5

Apply the distributive property.

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

Step 1.1.7.6

Multiply $x$ by $x$ by adding the exponents.

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Step 1.1.7.6.1

Move $x$ .

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

Step 1.1.7.6.2

Multiply $x$ by $x$ .

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

Step 1.1.8

Move $A$ .

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

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

$1 = A + B$

Step 1.2.2

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.

$0 = C$

Step 1.2.3

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 1.2.4

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

$1 = A + B$

$0 = C$

$- 1 = A$

$1 = A + B$

$0 = C$

$- 1 = A$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Rewrite the equation as $C = 0$ .

$C = 0$

$1 = A + B$

$- 1 = A$

Step 1.3.2

Rewrite the equation as $A = - 1$ .

$A = - 1$

$C = 0$

$1 = A + B$

Step 1.3.3

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

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Step 1.3.3.1

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

$1 = (- 1) + B$

$A = - 1$

$C = 0$

Step 1.3.3.2

Simplify the right side.

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Step 1.3.3.2.1

Remove parentheses.

$1 = - 1 + B$

$A = - 1$

$C = 0$

$1 = - 1 + B$

$A = - 1$

$C = 0$

$1 = - 1 + B$

$A = - 1$

$C = 0$

Step 1.3.4

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

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Step 1.3.4.1

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

$- 1 + B = 1$

$A = - 1$

$C = 0$

Step 1.3.4.2

Move all terms not containing $B$ to the right side of the equation.

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Step 1.3.4.2.1

Add $1$ to both sides of the equation.

$B = 1 + 1$

$A = - 1$

$C = 0$

Step 1.3.4.2.2

Add $1$ and $1$ .

$B = 2$

$A = - 1$

$C = 0$

$B = 2$

$A = - 1$

$C = 0$

$B = 2$

$A = - 1$

$C = 0$

Step 1.3.5

Solve the system of equations.

$B = 2 A = - 1 C = 0$

Step 1.3.6

List all of the solutions.

$B = 2, A = - 1, C = 0$

Step 1.4

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

$- \frac{1}{x} + \frac{2 x + 0}{x^{2} + 1}$

Step 1.5

Simplify.

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Step 1.5.1

Remove parentheses.

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

Step 1.5.2

Add $2 x$ and $0$ .

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

Step 1.5.3

Move the negative in front of the fraction.

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

Step 2

Split the single integral into multiple integrals.

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

Step 3

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

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

Step 4

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

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

Step 5

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

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

Step 6

Let $u = x^{2} + 1$ . Then $d u = 2 x d x$ , so $\frac{1}{2} d u = x d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.1

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

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Step 6.1.1

Differentiate $x^{2} + 1$ .

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

Step 6.1.2

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

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

Step 6.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

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

Step 6.1.4

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

$2 x + 0$

Step 6.1.5

Add $2 x$ and $0$ .

$2 x$

Step 6.2

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

$- (\ln (| x |) + C) + 2 \int \frac{1}{u} \cdot \frac{1}{2} d u$

Step 7

Simplify.

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Step 7.1

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

$- (\ln (| x |) + C) + 2 \int \frac{1}{u \cdot 2} d u$

Step 7.2

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

$- (\ln (| x |) + C) + 2 \int \frac{1}{2 u} d u$

Step 8

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$- (\ln (| x |) + C) + 2 (\frac{1}{2} \int \frac{1}{u} d u)$

Step 9

Simplify.

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Step 9.1

Combine $\frac{1}{2}$ and $2$ .

$- (\ln (| x |) + C) + \frac{2}{2} \int \frac{1}{u} d u$

Step 9.2

Cancel the common factor of $2$ .

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Step 9.2.1

Cancel the common factor.

$- (\ln (| x |) + C) + \frac{2}{2} \int \frac{1}{u} d u$

Step 9.2.2

Rewrite the expression.

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

Step 9.3

Multiply $\int \frac{1}{u} d u$ by $1$ .

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

Step 10

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

$- (\ln (| x |) + C) + \ln (| u |) + C$

Step 11

Simplify.

$- \ln (| x |) + \ln (| u |) + C$

Step 12

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

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