Calculus Examples

$\int \frac{2 x^{2} - 5 x + 2}{x^{3} + x} d x$

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

Factor by grouping.

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Step 1.1.1.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 2 = 4$ and whose sum is $b = - 5$ .

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Step 1.1.1.1.1.1

Factor $- 5$ out of $- 5 x$ .

$\frac{2 x^{2} - 5 (x) + 2}{x^{3} + x}$

Step 1.1.1.1.1.2

Rewrite $- 5$ as $- 1$ plus $- 4$

$\frac{2 x^{2} + (- 1 - 4) x + 2}{x^{3} + x}$

Step 1.1.1.1.1.3

Apply the distributive property.

$\frac{2 x^{2} - 1 x - 4 x + 2}{x^{3} + x}$

Step 1.1.1.1.2

Factor out the greatest common factor from each group.

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Step 1.1.1.1.2.1

Group the first two terms and the last two terms.

$\frac{(2 x^{2} - 1 x) - 4 x + 2}{x^{3} + x}$

Step 1.1.1.1.2.2

Factor out the greatest common factor (GCF) from each group.

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

Step 1.1.1.1.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

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

Step 1.1.1.2

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

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Step 1.1.1.2.1

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

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

Step 1.1.1.2.2

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

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

Step 1.1.1.2.3

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

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

Step 1.1.1.2.4

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

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

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 $x (x^{2} + 1)$ .

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

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.

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

Step 1.1.4.1.2

Rewrite the expression.

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

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.

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

Step 1.1.4.2.2

Divide $(2 x - 1) (x - 2)$ by $1$ .

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

Step 1.1.5

Expand $(2 x - 1) (x - 2)$ using the FOIL Method.

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Step 1.1.5.1

Apply the distributive property.

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

Step 1.1.5.2

Apply the distributive property.

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

Step 1.1.5.3

Apply the distributive property.

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

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 $x$ by adding the exponents.

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Step 1.1.6.1.1.1

Move $x$ .

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

Step 1.1.6.1.1.2

Multiply $x$ by $x$ .

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

Step 1.1.6.1.2

Multiply $- 2$ by $2$ .

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

Step 1.1.6.1.3

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

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

Step 1.1.6.1.4

Multiply $- 1$ by $- 2$ .

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

Step 1.1.6.2

Subtract $x$ from $- 4 x$ .

$2 x^{2} - 5 x + 2 = \frac{A (x (x^{2} + 1))}{x} + \frac{(B x + C) (x (x^{2} + 1))}{x^{2} + 1}$

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.

$2 x^{2} - 5 x + 2 = \frac{A (x (x^{2} + 1))}{x} + \frac{(B x + C) (x (x^{2} + 1))}{x^{2} + 1}$

Step 1.1.7.1.2

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

$2 x^{2} - 5 x + 2 = A (x^{2} + 1) + \frac{(B x + C) (x (x^{2} + 1))}{x^{2} + 1}$

Step 1.1.7.2

Apply the distributive property.

$2 x^{2} - 5 x + 2 = A x^{2} + A \cdot 1 + \frac{(B x + C) (x (x^{2} + 1))}{x^{2} + 1}$

Step 1.1.7.3

Multiply $A$ by $1$ .

$2 x^{2} - 5 x + 2 = A x^{2} + A + \frac{(B x + C) (x (x^{2} + 1))}{x^{2} + 1}$

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.

$2 x^{2} - 5 x + 2 = A x^{2} + A + \frac{(B x + C) (x (x^{2} + 1))}{x^{2} + 1}$

Step 1.1.7.4.2

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

$2 x^{2} - 5 x + 2 = A x^{2} + A + (B x + C) (x)$

Step 1.1.7.5

Apply the distributive property.

$2 x^{2} - 5 x + 2 = A x^{2} + A + B x \cdot x + C x$

Step 1.1.7.6

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

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Step 1.1.7.6.1

Move $x$ .

$2 x^{2} - 5 x + 2 = A x^{2} + A + B (x \cdot x) + C x$

Step 1.1.7.6.2

Multiply $x$ by $x$ .

$2 x^{2} - 5 x + 2 = A x^{2} + A + B x^{2} + C x$

Step 1.1.8

Move $A$ .

$2 x^{2} - 5 x + 2 = A x^{2} + B x^{2} + C x + A$

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.

$2 = A + B$

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.

$- 5 = C$

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.

$2 = A$

Step 1.2.4

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

$2 = A + B$

$- 5 = C$

$2 = A$

$2 = A + B$

$- 5 = C$

$2 = A$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Rewrite the equation as $C = - 5$ .

$C = - 5$

$2 = A + B$

$2 = A$

Step 1.3.2

Rewrite the equation as $A = 2$ .

$A = 2$

$C = - 5$

$2 = A + B$

Step 1.3.3

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

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Step 1.3.3.1

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

$2 = (2) + B$

$A = 2$

$C = - 5$

Step 1.3.3.2

Simplify the right side.

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Step 1.3.3.2.1

Remove parentheses.

$2 = 2 + B$

$A = 2$

$C = - 5$

$2 = 2 + B$

$A = 2$

$C = - 5$

$2 = 2 + B$

$A = 2$

$C = - 5$

Step 1.3.4

Solve for $B$ in $2 = 2 + B$ .

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Step 1.3.4.1

Rewrite the equation as $2 + B = 2$ .

$2 + B = 2$

$A = 2$

$C = - 5$

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

Subtract $2$ from both sides of the equation.

$B = 2 - 2$

$A = 2$

$C = - 5$

Step 1.3.4.2.2

Subtract $2$ from $2$ .

$B = 0$

$A = 2$

$C = - 5$

$B = 0$

$A = 2$

$C = - 5$

$B = 0$

$A = 2$

$C = - 5$

Step 1.3.5

Solve the system of equations.

$B = 0 A = 2 C = - 5$

Step 1.3.6

List all of the solutions.

$B = 0, A = 2, C = - 5$

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 $C$ .

$\frac{2}{x} + \frac{0 x - 5}{x^{2} + 1}$

Step 1.5

Simplify.

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Step 1.5.1

Remove parentheses.

$\frac{2}{x} + \frac{(0) \cdot x - 5}{x^{2} + 1}$

Step 1.5.2

Simplify the numerator.

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Step 1.5.2.1

Factor $5$ out of $0 \cdot x - 5$ .

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Step 1.5.2.1.1

Factor $5$ out of $0 \cdot x$ .

$\frac{2}{x} + \frac{5 (0 \cdot x) - 5}{x^{2} + 1}$

Step 1.5.2.1.2

Factor $5$ out of $- 5$ .

$\frac{2}{x} + \frac{5 (0 \cdot x) + 5 (- 1)}{x^{2} + 1}$

Step 1.5.2.1.3

Factor $5$ out of $5 (0 \cdot x) + 5 (- 1)$ .

$\frac{2}{x} + \frac{5 (0 \cdot x - 1)}{x^{2} + 1}$

Step 1.5.2.2

Multiply $0$ by $x$ .

$\frac{2}{x} + \frac{5 (0 - 1)}{x^{2} + 1}$

Step 1.5.2.3

Subtract $1$ from $0$ .

$\frac{2}{x} + \frac{5 \cdot - 1}{x^{2} + 1}$

Step 1.5.3

Simplify the expression.

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Step 1.5.3.1

Multiply $5$ by $- 1$ .

$\frac{2}{x} + \frac{- 5}{x^{2} + 1}$

Step 1.5.3.2

Move the negative in front of the fraction.

$\int \frac{2}{x} - \frac{5}{x^{2} + 1} d x$

Step 2

Split the single integral into multiple integrals.

$\int \frac{2}{x} d x + \int - \frac{5}{x^{2} + 1} d x$

Step 3

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

$2 \int \frac{1}{x} d x + \int - \frac{5}{x^{2} + 1} d x$

Step 4

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

$2 (\ln (| x |) + C) + \int - \frac{5}{x^{2} + 1} d x$

Step 5

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

$2 (\ln (| x |) + C) - \int \frac{5}{x^{2} + 1} d x$

Step 6

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

$2 (\ln (| x |) + C) - (5 \int \frac{1}{x^{2} + 1} d x)$

Step 7

Simplify the expression.

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Step 7.1

Multiply $5$ by $- 1$ .

$2 (\ln (| x |) + C) - 5 \int \frac{1}{x^{2} + 1} d x$

Step 7.2

Reorder $x^{2}$ and $1$ .

$2 (\ln (| x |) + C) - 5 \int \frac{1}{1 + x^{2}} d x$

Step 7.3

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

$2 (\ln (| x |) + C) - 5 \int \frac{1}{1^{2} + x^{2}} d x$

Step 8

The integral of $\frac{1}{1^{2} + x^{2}}$ with respect to $x$ is $\arctan (x) + C$ .

$2 (\ln (| x |) + C) - 5 (\arctan (x) + C)$

Step 9

Simplify.

$2 \ln (| x |) - 5 \arctan (x) + C$