Calculus Examples

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

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 $x$ out of $x^{3} + 4 x$ .

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Step 1.1.1.1

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

$\frac{2 x^{2} - x + 4}{x \cdot x^{2} + 4 x}$

Step 1.1.1.2

Factor $x$ out of $4 x$ .

$\frac{2 x^{2} - x + 4}{x \cdot x^{2} + x \cdot 4}$

Step 1.1.1.3

Factor $x$ out of $x \cdot x^{2} + x \cdot 4$ .

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

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.

$\frac{A}{x} + \frac{B x + C}{x^{2} + 4}$

Step 1.1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $x (x^{2} + 4)$ .

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

Step 1.1.4

Cancel the common factor of $x$ .

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Step 1.1.4.1

Cancel the common factor.

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

Step 1.1.4.2

Rewrite the expression.

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

Step 1.1.5

Cancel the common factor of $x^{2} + 4$ .

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Step 1.1.5.1

Cancel the common factor.

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

Step 1.1.5.2

Divide $2 x^{2} - x + 4$ by $1$ .

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

Step 1.1.6

Simplify each term.

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Step 1.1.6.1

Cancel the common factor of $x$ .

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Step 1.1.6.1.1

Cancel the common factor.

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

Step 1.1.6.1.2

Divide $A (x^{2} + 4)$ by $1$ .

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

Step 1.1.6.2

Apply the distributive property.

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

Step 1.1.6.3

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

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

Step 1.1.6.4

Cancel the common factor of $x^{2} + 4$ .

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Step 1.1.6.4.1

Cancel the common factor.

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

Step 1.1.6.4.2

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

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

Step 1.1.6.5

Apply the distributive property.

$2 x^{2} - x + 4 = A x^{2} + 4 A + B x \cdot x + C x$

Step 1.1.6.6

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

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Step 1.1.6.6.1

Move $x$ .

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

Step 1.1.6.6.2

Multiply $x$ by $x$ .

$2 x^{2} - x + 4 = A x^{2} + 4 A + B x^{2} + C x$

Step 1.1.7

Move $4 A$ .

$2 x^{2} - x + 4 = A x^{2} + B x^{2} + C x + 4 A$

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.

$- 1 = C$

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.

$4 = 4 A$

Step 1.2.4

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

$2 = A + B$

$- 1 = C$

$4 = 4 A$

$2 = A + B$

$- 1 = C$

$4 = 4 A$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Rewrite the equation as $C = - 1$ .

$C = - 1$

$2 = A + B$

$4 = 4 A$

Step 1.3.2

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

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Step 1.3.2.1

Rewrite the equation as $4 A = 4$ .

$4 A = 4$

$C = - 1$

$2 = A + B$

Step 1.3.2.2

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

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Step 1.3.2.2.1

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

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

$C = - 1$

$2 = A + B$

Step 1.3.2.2.2

Simplify the left side.

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Step 1.3.2.2.2.1

Cancel the common factor of $4$ .

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Step 1.3.2.2.2.1.1

Cancel the common factor.

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

$C = - 1$

$2 = A + B$

Step 1.3.2.2.2.1.2

Divide $A$ by $1$ .

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

$C = - 1$

$2 = A + B$

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

$C = - 1$

$2 = A + B$

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

$C = - 1$

$2 = A + B$

Step 1.3.2.2.3

Simplify the right side.

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Step 1.3.2.2.3.1

Divide $4$ by $4$ .

$A = 1$

$C = - 1$

$2 = A + B$

$A = 1$

$C = - 1$

$2 = A + B$

$A = 1$

$C = - 1$

$2 = A + B$

$A = 1$

$C = - 1$

$2 = A + B$

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 $2 = A + B$ with $1$ .

$2 = (1) + B$

$A = 1$

$C = - 1$

Step 1.3.3.2

Simplify the right side.

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Step 1.3.3.2.1

Remove parentheses.

$2 = 1 + B$

$A = 1$

$C = - 1$

$2 = 1 + B$

$A = 1$

$C = - 1$

$2 = 1 + B$

$A = 1$

$C = - 1$

Step 1.3.4

Solve for $B$ in $2 = 1 + B$ .

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Step 1.3.4.1

Rewrite the equation as $1 + B = 2$ .

$1 + B = 2$

$A = 1$

$C = - 1$

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 $1$ from both sides of the equation.

$B = 2 - 1$

$A = 1$

$C = - 1$

Step 1.3.4.2.2

Subtract $1$ from $2$ .

$B = 1$

$A = 1$

$C = - 1$

$B = 1$

$A = 1$

$C = - 1$

$B = 1$

$A = 1$

$C = - 1$

Step 1.3.5

Solve the system of equations.

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

Step 1.3.6

List all of the solutions.

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

Step 1.4

Replace each of the partial fraction coefficients in $\frac{A}{x} + \frac{B x + C}{x^{2} + 4}$ with the values found for $A$ , $B$ , and $C$ .

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

Step 1.5

Simplify.

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Step 1.5.1

Remove parentheses.

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

Step 1.5.2

Multiply $x$ by $1$ .

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

Step 2

Split the single integral into multiple integrals.

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

Step 3

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

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

Step 4

Split the fraction $\frac{x - 1}{x^{2} + 4}$ into two fractions.

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

Step 5

Move the negative in front of the fraction.

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

Step 6

Split the single integral into multiple integrals.

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

Step 7

Let $u = x^{2} + 4$ . 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 7.1

Let $u = x^{2} + 4$ . Find $\frac{d u}{d x}$ .

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Step 7.1.1

Differentiate $x^{2} + 4$ .

$\frac{d}{d x} [x^{2} + 4]$

Step 7.1.2

By the Sum Rule, the derivative of $x^{2} + 4$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [4]$ .

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [4]$

Step 7.1.3

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

$2 x + \frac{d}{d x} [4]$

Step 7.1.4

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$2 x + 0$

Step 7.1.5

Add $2 x$ and $0$ .

$2 x$

Step 7.2

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

$\ln (| x |) + C + \int \frac{1}{u} \cdot \frac{1}{2} d u + \int - \frac{1}{x^{2} + 4} d x$

Step 8

Simplify.

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Step 8.1

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

$\ln (| x |) + C + \int \frac{1}{u \cdot 2} d u + \int - \frac{1}{x^{2} + 4} d x$

Step 8.2

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

$\ln (| x |) + C + \int \frac{1}{2 u} d u + \int - \frac{1}{x^{2} + 4} d x$

Step 9

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

$\ln (| x |) + C + \frac{1}{2} \int \frac{1}{u} d u + \int - \frac{1}{x^{2} + 4} d x$

Step 10

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

$\ln (| x |) + C + \frac{1}{2} (\ln (| u |) + C) + \int - \frac{1}{x^{2} + 4} d x$

Step 11

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

$\ln (| x |) + C + \frac{1}{2} (\ln (| u |) + C) - \int \frac{1}{x^{2} + 4} d x$

Step 12

Simplify the expression.

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Step 12.1

Reorder $x^{2}$ and $4$ .

$\ln (| x |) + C + \frac{1}{2} (\ln (| u |) + C) - \int \frac{1}{4 + x^{2}} d x$

Step 12.2

Rewrite $4$ as $2^{2}$ .

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

Step 13

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

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

Step 14

Simplify.

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Step 14.1

Combine $\frac{1}{2}$ and $\arctan (\frac{x}{2})$ .

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

Step 14.2

Simplify.

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

Step 15

Replace all occurrences of $u$ with $x^{2} + 4$ .

$\ln (| x |) + \frac{1}{2} \ln (| x^{2} + 4 |) - \frac{1}{2} \arctan (\frac{1}{2} x) + C$