Calculus Examples

$\int \frac{24 x^{3} + 20 x}{3 x^{4} + 5 x^{2}} d x$

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 $4 x$ out of $24 x^{3} + 20 x$ .

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Step 1.1.1.1.1

Factor $4 x$ out of $24 x^{3}$ .

$\frac{4 x (6 x^{2}) + 20 x}{3 x^{4} + 5 x^{2}}$

Step 1.1.1.1.2

Factor $4 x$ out of $20 x$ .

$\frac{4 x (6 x^{2}) + 4 x (5)}{3 x^{4} + 5 x^{2}}$

Step 1.1.1.1.3

Factor $4 x$ out of $4 x (6 x^{2}) + 4 x (5)$ .

$\frac{4 x (6 x^{2} + 5)}{3 x^{4} + 5 x^{2}}$

Step 1.1.1.2

Factor $x^{2}$ out of $3 x^{4} + 5 x^{2}$ .

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Step 1.1.1.2.1

Factor $x^{2}$ out of $3 x^{4}$ .

$\frac{4 x (6 x^{2} + 5)}{x^{2} (3 x^{2}) + 5 x^{2}}$

Step 1.1.1.2.2

Factor $x^{2}$ out of $5 x^{2}$ .

$\frac{4 x (6 x^{2} + 5)}{x^{2} (3 x^{2}) + x^{2} \cdot 5}$

Step 1.1.1.2.3

Factor $x^{2}$ out of $x^{2} (3 x^{2}) + x^{2} \cdot 5$ .

$\frac{4 x (6 x^{2} + 5)}{x^{2} (3 x^{2} + 5)}$

Step 1.1.1.3

Reduce the expression $\frac{4 x (6 x^{2} + 5)}{x^{2} (3 x^{2} + 5)}$ by cancelling the common factors.

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Step 1.1.1.3.1

Factor $x$ out of $4 x (6 x^{2} + 5)$ .

$\frac{x (4 (6 x^{2} + 5))}{x^{2} (3 x^{2} + 5)}$

Step 1.1.1.3.2

Factor $x$ out of $x^{2} (3 x^{2} + 5)$ .

$\frac{x (4 (6 x^{2} + 5))}{x (x (3 x^{2} + 5))}$

Step 1.1.1.3.3

Cancel the common factor.

$\frac{x (4 (6 x^{2} + 5))}{x (x (3 x^{2} + 5))}$

Step 1.1.1.3.4

Rewrite the expression.

$\frac{4 (6 x^{2} + 5)}{x (3 x^{2} + 5)}$

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.

$\frac{A}{x} + \frac{B x + C}{3 x^{2} + 5}$

Step 1.1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $x (3 x^{2} + 5)$ .

$\frac{4 (6 x^{2} + 5) (x (3 x^{2} + 5))}{x (3 x^{2} + 5)} = \frac{A (x (3 x^{2} + 5))}{x} + \frac{(B x + C) (x (3 x^{2} + 5))}{3 x^{2} + 5}$

Step 1.1.4

Cancel the common factor of $x$ .

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Step 1.1.4.1

Cancel the common factor.

$\frac{4 (6 x^{2} + 5) (x (3 x^{2} + 5))}{x (3 x^{2} + 5)} = \frac{A (x (3 x^{2} + 5))}{x} + \frac{(B x + C) (x (3 x^{2} + 5))}{3 x^{2} + 5}$

Step 1.1.4.2

Rewrite the expression.

$\frac{4 (6 x^{2} + 5) (3 x^{2} + 5)}{3 x^{2} + 5} = \frac{A (x (3 x^{2} + 5))}{x} + \frac{(B x + C) (x (3 x^{2} + 5))}{3 x^{2} + 5}$

Step 1.1.5

Cancel the common factor of $3 x^{2} + 5$ .

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Step 1.1.5.1

Cancel the common factor.

$\frac{4 (6 x^{2} + 5) (3 x^{2} + 5)}{3 x^{2} + 5} = \frac{A (x (3 x^{2} + 5))}{x} + \frac{(B x + C) (x (3 x^{2} + 5))}{3 x^{2} + 5}$

Step 1.1.5.2

Divide $4 (6 x^{2} + 5)$ by $1$ .

$4 (6 x^{2} + 5) = \frac{A (x (3 x^{2} + 5))}{x} + \frac{(B x + C) (x (3 x^{2} + 5))}{3 x^{2} + 5}$

Step 1.1.6

Apply the distributive property.

$4 (6 x^{2}) + 4 \cdot 5 = \frac{A (x (3 x^{2} + 5))}{x} + \frac{(B x + C) (x (3 x^{2} + 5))}{3 x^{2} + 5}$

Step 1.1.7

Multiply.

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Step 1.1.7.1

Multiply $6$ by $4$ .

$24 x^{2} + 4 \cdot 5 = \frac{A (x (3 x^{2} + 5))}{x} + \frac{(B x + C) (x (3 x^{2} + 5))}{3 x^{2} + 5}$

Step 1.1.7.2

Multiply $4$ by $5$ .

$24 x^{2} + 20 = \frac{A (x (3 x^{2} + 5))}{x} + \frac{(B x + C) (x (3 x^{2} + 5))}{3 x^{2} + 5}$

Step 1.1.8

Simplify each term.

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Step 1.1.8.1

Cancel the common factor of $x$ .

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Step 1.1.8.1.1

Cancel the common factor.

$24 x^{2} + 20 = \frac{A (x (3 x^{2} + 5))}{x} + \frac{(B x + C) (x (3 x^{2} + 5))}{3 x^{2} + 5}$

Step 1.1.8.1.2

Divide $A (3 x^{2} + 5)$ by $1$ .

$24 x^{2} + 20 = A (3 x^{2} + 5) + \frac{(B x + C) (x (3 x^{2} + 5))}{3 x^{2} + 5}$

Step 1.1.8.2

Apply the distributive property.

$24 x^{2} + 20 = A (3 x^{2}) + A \cdot 5 + \frac{(B x + C) (x (3 x^{2} + 5))}{3 x^{2} + 5}$

Step 1.1.8.3

Rewrite using the commutative property of multiplication.

$24 x^{2} + 20 = 3 A x^{2} + A \cdot 5 + \frac{(B x + C) (x (3 x^{2} + 5))}{3 x^{2} + 5}$

Step 1.1.8.4

Move $5$ to the left of $A$ .

$24 x^{2} + 20 = 3 A x^{2} + 5 \cdot A + \frac{(B x + C) (x (3 x^{2} + 5))}{3 x^{2} + 5}$

Step 1.1.8.5

Cancel the common factor of $3 x^{2} + 5$ .

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Step 1.1.8.5.1

Cancel the common factor.

$24 x^{2} + 20 = 3 A x^{2} + 5 A + \frac{(B x + C) (x (3 x^{2} + 5))}{3 x^{2} + 5}$

Step 1.1.8.5.2

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

$24 x^{2} + 20 = 3 A x^{2} + 5 A + (B x + C) (x)$

Step 1.1.8.6

Apply the distributive property.

$24 x^{2} + 20 = 3 A x^{2} + 5 A + B x \cdot x + C x$

Step 1.1.8.7

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

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Step 1.1.8.7.1

Move $x$ .

$24 x^{2} + 20 = 3 A x^{2} + 5 A + B (x \cdot x) + C x$

Step 1.1.8.7.2

Multiply $x$ by $x$ .

$24 x^{2} + 20 = 3 A x^{2} + 5 A + B x^{2} + C x$

Step 1.1.9

Simplify the expression.

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Step 1.1.9.1

Move $A$ .

$24 x^{2} + 20 = 3 x^{2} A + 5 A + B x^{2} + C x$

Step 1.1.9.2

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

$24 x^{2} + 20 = 3 x^{2} A + 5 A + B x^{2} + C x$

Step 1.1.9.3

Move $5 A$ .

$24 x^{2} + 20 = 3 x^{2} A + B x^{2} + C x + 5 A$

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.

$24 = 3 A + B$

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.

$20 = 5 A$

Step 1.2.4

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

$24 = 3 A + B$

$0 = C$

$20 = 5 A$

$24 = 3 A + B$

$0 = C$

$20 = 5 A$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Rewrite the equation as $C = 0$ .

$C = 0$

$24 = 3 A + B$

$20 = 5 A$

Step 1.3.2

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

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Step 1.3.2.1

Rewrite the equation as $5 A = 20$ .

$5 A = 20$

$C = 0$

$24 = 3 A + B$

Step 1.3.2.2

Divide each term in $5 A = 20$ by $5$ and simplify.

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Step 1.3.2.2.1

Divide each term in $5 A = 20$ by $5$ .

$\frac{5 A}{5} = \frac{20}{5}$

$C = 0$

$24 = 3 A + B$

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 $5$ .

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Step 1.3.2.2.2.1.1

Cancel the common factor.

$\frac{5 A}{5} = \frac{20}{5}$

$C = 0$

$24 = 3 A + B$

Step 1.3.2.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{20}{5}$

$C = 0$

$24 = 3 A + B$

$A = \frac{20}{5}$

$C = 0$

$24 = 3 A + B$

$A = \frac{20}{5}$

$C = 0$

$24 = 3 A + B$

Step 1.3.2.2.3

Simplify the right side.

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Step 1.3.2.2.3.1

Divide $20$ by $5$ .

$A = 4$

$C = 0$

$24 = 3 A + B$

$A = 4$

$C = 0$

$24 = 3 A + B$

$A = 4$

$C = 0$

$24 = 3 A + B$

$A = 4$

$C = 0$

$24 = 3 A + B$

Step 1.3.3

Replace all occurrences of $A$ with $4$ in each equation.

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Step 1.3.3.1

Replace all occurrences of $A$ in $24 = 3 A + B$ with $4$ .

$24 = 3 (4) + B$

$A = 4$

$C = 0$

Step 1.3.3.2

Simplify the right side.

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Step 1.3.3.2.1

Multiply $3$ by $4$ .

$24 = 12 + B$

$A = 4$

$C = 0$

$24 = 12 + B$

$A = 4$

$C = 0$

$24 = 12 + B$

$A = 4$

$C = 0$

Step 1.3.4

Solve for $B$ in $24 = 12 + B$ .

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Step 1.3.4.1

Rewrite the equation as $12 + B = 24$ .

$12 + B = 24$

$A = 4$

$C = 0$

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

Subtract $12$ from both sides of the equation.

$B = 24 - 12$

$A = 4$

$C = 0$

Step 1.3.4.2.2

Subtract $12$ from $24$ .

$B = 12$

$A = 4$

$C = 0$

$B = 12$

$A = 4$

$C = 0$

$B = 12$

$A = 4$

$C = 0$

Step 1.3.5

Solve the system of equations.

$B = 12 A = 4 C = 0$

Step 1.3.6

List all of the solutions.

$B = 12, A = 4, C = 0$

Step 1.4

Replace each of the partial fraction coefficients in $\frac{A}{x} + \frac{B x + C}{3 x^{2} + 5}$ with the values found for $A$ , $B$ , and $C$ .

$\frac{4}{x} + \frac{12 x + 0}{3 x^{2} + 5}$

Step 1.5

Simplify.

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Step 1.5.1

Remove parentheses.

$\frac{4}{x} + \frac{(12) \cdot x + 0}{3 x^{2} + 5}$

Step 1.5.2

Add $12 x$ and $0$ .

$\int \frac{4}{x} + \frac{12 x}{3 x^{2} + 5} d x$

Step 2

Split the single integral into multiple integrals.

$\int \frac{4}{x} d x + \int \frac{12 x}{3 x^{2} + 5} d x$

Step 3

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

$4 \int \frac{1}{x} d x + \int \frac{12 x}{3 x^{2} + 5} d x$

Step 4

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

$4 (\ln (| x |) + C) + \int \frac{12 x}{3 x^{2} + 5} d x$

Step 5

Since $12$ is constant with respect to $x$ , move $12$ out of the integral.

$4 (\ln (| x |) + C) + 12 \int \frac{x}{3 x^{2} + 5} d x$

Step 6

Let $u = 3 x^{2} + 5$ . Then $d u = 6 x d x$ , so $\frac{1}{6} d u = x d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.1

Let $u = 3 x^{2} + 5$ . Find $\frac{d u}{d x}$ .

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Step 6.1.1

Differentiate $3 x^{2} + 5$ .

$\frac{d}{d x} [3 x^{2} + 5]$

Step 6.1.2

By the Sum Rule, the derivative of $3 x^{2} + 5$ with respect to $x$ is $\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [5]$ .

$\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [5]$

Step 6.1.3

Evaluate $\frac{d}{d x} [3 x^{2}]$ .

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Step 6.1.3.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{2}$ with respect to $x$ is $3 \frac{d}{d x} [x^{2}]$ .

$3 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [5]$

Step 6.1.3.2

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

$3 (2 x) + \frac{d}{d x} [5]$

Step 6.1.3.3

Multiply $2$ by $3$ .

$6 x + \frac{d}{d x} [5]$

Step 6.1.4

Differentiate using the Constant Rule.

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Step 6.1.4.1

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$6 x + 0$

Step 6.1.4.2

Add $6 x$ and $0$ .

$6 x$

Step 6.2

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

$4 (\ln (| x |) + C) + 12 \int \frac{1}{u} \cdot \frac{1}{6} d u$

Step 7

Simplify.

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Step 7.1

Multiply $\frac{1}{u}$ by $\frac{1}{6}$ .

$4 (\ln (| x |) + C) + 12 \int \frac{1}{u \cdot 6} d u$

Step 7.2

Move $6$ to the left of $u$ .

$4 (\ln (| x |) + C) + 12 \int \frac{1}{6 u} d u$

Step 8

Since $\frac{1}{6}$ is constant with respect to $u$ , move $\frac{1}{6}$ out of the integral.

$4 (\ln (| x |) + C) + 12 (\frac{1}{6} \int \frac{1}{u} d u)$

Step 9

Simplify.

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Step 9.1

Combine $\frac{1}{6}$ and $12$ .

$4 (\ln (| x |) + C) + \frac{12}{6} \int \frac{1}{u} d u$

Step 9.2

Cancel the common factor of $12$ and $6$ .

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Step 9.2.1

Factor $6$ out of $12$ .

$4 (\ln (| x |) + C) + \frac{6 \cdot 2}{6} \int \frac{1}{u} d u$

Step 9.2.2

Cancel the common factors.

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Step 9.2.2.1

Factor $6$ out of $6$ .

$4 (\ln (| x |) + C) + \frac{6 \cdot 2}{6 (1)} \int \frac{1}{u} d u$

Step 9.2.2.2

Cancel the common factor.

$4 (\ln (| x |) + C) + \frac{6 \cdot 2}{6 \cdot 1} \int \frac{1}{u} d u$

Step 9.2.2.3

Rewrite the expression.

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

Step 9.2.2.4

Divide $2$ by $1$ .

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

Step 10

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

$4 (\ln (| x |) + C) + 2 (\ln (| u |) + C)$

Step 11

Simplify.

$4 \ln (| x |) + 2 \ln (| u |) + C$

Step 12

Replace all occurrences of $u$ with $3 x^{2} + 5$ .

$4 \ln (| x |) + 2 \ln (| 3 x^{2} + 5 |) + C$