Calculus Examples

$\int \frac{w^{2} - w + 6}{w^{3} + 3 w} d w$

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 $w$ out of $w^{3} + 3 w$ .

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Step 1.1.1.1

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

$\frac{w^{2} - w + 6}{w \cdot w^{2} + 3 w}$

Step 1.1.1.2

Factor $w$ out of $3 w$ .

$\frac{w^{2} - w + 6}{w \cdot w^{2} + w \cdot 3}$

Step 1.1.1.3

Factor $w$ out of $w \cdot w^{2} + w \cdot 3$ .

$\frac{w^{2} - w + 6}{w (w^{2} + 3)}$

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.

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

Step 1.1.3

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

$\frac{(w^{2} - w + 6) (w (w^{2} + 3))}{w (w^{2} + 3)} = \frac{A (w (w^{2} + 3))}{w} + \frac{(B w + C) (w (w^{2} + 3))}{w^{2} + 3}$

Step 1.1.4

Cancel the common factor of $w$ .

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Step 1.1.4.1

Cancel the common factor.

$\frac{(w^{2} - w + 6) (w (w^{2} + 3))}{w (w^{2} + 3)} = \frac{A (w (w^{2} + 3))}{w} + \frac{(B w + C) (w (w^{2} + 3))}{w^{2} + 3}$

Step 1.1.4.2

Rewrite the expression.

$\frac{(w^{2} - w + 6) (w^{2} + 3)}{w^{2} + 3} = \frac{A (w (w^{2} + 3))}{w} + \frac{(B w + C) (w (w^{2} + 3))}{w^{2} + 3}$

Step 1.1.5

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

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Step 1.1.5.1

Cancel the common factor.

$\frac{(w^{2} - w + 6) (w^{2} + 3)}{w^{2} + 3} = \frac{A (w (w^{2} + 3))}{w} + \frac{(B w + C) (w (w^{2} + 3))}{w^{2} + 3}$

Step 1.1.5.2

Divide $w^{2} - w + 6$ by $1$ .

$w^{2} - w + 6 = \frac{A (w (w^{2} + 3))}{w} + \frac{(B w + C) (w (w^{2} + 3))}{w^{2} + 3}$

Step 1.1.6

Simplify each term.

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Step 1.1.6.1

Cancel the common factor of $w$ .

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Step 1.1.6.1.1

Cancel the common factor.

$w^{2} - w + 6 = \frac{A (w (w^{2} + 3))}{w} + \frac{(B w + C) (w (w^{2} + 3))}{w^{2} + 3}$

Step 1.1.6.1.2

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

$w^{2} - w + 6 = A (w^{2} + 3) + \frac{(B w + C) (w (w^{2} + 3))}{w^{2} + 3}$

Step 1.1.6.2

Apply the distributive property.

$w^{2} - w + 6 = A w^{2} + A \cdot 3 + \frac{(B w + C) (w (w^{2} + 3))}{w^{2} + 3}$

Step 1.1.6.3

Move $3$ to the left of $A$ .

$w^{2} - w + 6 = A w^{2} + 3 \cdot A + \frac{(B w + C) (w (w^{2} + 3))}{w^{2} + 3}$

Step 1.1.6.4

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

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Step 1.1.6.4.1

Cancel the common factor.

$w^{2} - w + 6 = A w^{2} + 3 A + \frac{(B w + C) (w (w^{2} + 3))}{w^{2} + 3}$

Step 1.1.6.4.2

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

$w^{2} - w + 6 = A w^{2} + 3 A + (B w + C) (w)$

Step 1.1.6.5

Apply the distributive property.

$w^{2} - w + 6 = A w^{2} + 3 A + B w \cdot w + C w$

Step 1.1.6.6

Multiply $w$ by $w$ by adding the exponents.

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Step 1.1.6.6.1

Move $w$ .

$w^{2} - w + 6 = A w^{2} + 3 A + B (w \cdot w) + C w$

Step 1.1.6.6.2

Multiply $w$ by $w$ .

$w^{2} - w + 6 = A w^{2} + 3 A + B w^{2} + C w$

Step 1.1.7

Move $3 A$ .

$w^{2} - w + 6 = A w^{2} + B w^{2} + C w + 3 A$

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

$6 = 3 A$

Step 1.2.4

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

$1 = A + B$

$- 1 = C$

$6 = 3 A$

$1 = A + B$

$- 1 = C$

$6 = 3 A$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Rewrite the equation as $C = - 1$ .

$C = - 1$

$1 = A + B$

$6 = 3 A$

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 $3 A = 6$ .

$3 A = 6$

$C = - 1$

$1 = A + B$

Step 1.3.2.2

Divide each term in $3 A = 6$ by $3$ and simplify.

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Step 1.3.2.2.1

Divide each term in $3 A = 6$ by $3$ .

$\frac{3 A}{3} = \frac{6}{3}$

$C = - 1$

$1 = A + B$

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 $3$ .

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Step 1.3.2.2.2.1.1

Cancel the common factor.

$\frac{3 A}{3} = \frac{6}{3}$

$C = - 1$

$1 = A + B$

Step 1.3.2.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{6}{3}$

$C = - 1$

$1 = A + B$

$A = \frac{6}{3}$

$C = - 1$

$1 = A + B$

$A = \frac{6}{3}$

$C = - 1$

$1 = A + B$

Step 1.3.2.2.3

Simplify the right side.

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Step 1.3.2.2.3.1

Divide $6$ by $3$ .

$A = 2$

$C = - 1$

$1 = A + B$

$A = 2$

$C = - 1$

$1 = A + B$

$A = 2$

$C = - 1$

$1 = A + B$

$A = 2$

$C = - 1$

$1 = A + B$

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

$1 = (2) + B$

$A = 2$

$C = - 1$

Step 1.3.3.2

Simplify the right side.

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Step 1.3.3.2.1

Remove parentheses.

$1 = 2 + B$

$A = 2$

$C = - 1$

$1 = 2 + B$

$A = 2$

$C = - 1$

$1 = 2 + B$

$A = 2$

$C = - 1$

Step 1.3.4

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

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Step 1.3.4.1

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

$2 + B = 1$

$A = 2$

$C = - 1$

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 $2$ from both sides of the equation.

$B = 1 - 2$

$A = 2$

$C = - 1$

Step 1.3.4.2.2

Subtract $2$ from $1$ .

$B = - 1$

$A = 2$

$C = - 1$

$B = - 1$

$A = 2$

$C = - 1$

$B = - 1$

$A = 2$

$C = - 1$

Step 1.3.5

Solve the system of equations.

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

Step 1.3.6

List all of the solutions.

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

Step 1.4

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

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

Step 1.5

Simplify.

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Step 1.5.1

Remove parentheses.

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

Step 1.5.2

Factor $- 1$ out of $- 1 w$ .

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

Step 1.5.3

Rewrite $- 1$ as $- 1 (1)$ .

$\frac{2}{w} + \frac{- 1 w - 1 \cdot 1}{w^{2} + 3}$

Step 1.5.4

Factor $- 1$ out of $- 1 (w) - 1 (1)$ .

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

Step 1.5.5

Move the negative in front of the fraction.

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

Step 2

Split the single integral into multiple integrals.

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

Step 3

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

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

Step 4

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

$2 (\ln (| w |) + C) + \int - \frac{w + 1}{w^{2} + 3} d w$

Step 5

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

$2 (\ln (| w |) + C) - \int \frac{w + 1}{w^{2} + 3} d w$

Step 6

Split the fraction $\frac{w + 1}{w^{2} + 3}$ into two fractions.

$2 (\ln (| w |) + C) - \int \frac{w}{w^{2} + 3} + \frac{1}{w^{2} + 3} d w$

Step 7

Split the single integral into multiple integrals.

$2 (\ln (| w |) + C) - (\int \frac{w}{w^{2} + 3} d w + \int \frac{1}{w^{2} + 3} d w)$

Step 8

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

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Step 8.1

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

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Step 8.1.1

Differentiate $w^{2} + 3$ .

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

Step 8.1.2

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

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

Step 8.1.3

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

$2 w + \frac{d}{d w} [3]$

Step 8.1.4

Since $3$ is constant with respect to $w$ , the derivative of $3$ with respect to $w$ is $0$ .

$2 w + 0$

Step 8.1.5

Add $2 w$ and $0$ .

$2 w$

Step 8.2

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

$2 (\ln (| w |) + C) - (\int \frac{1}{u} \cdot \frac{1}{2} d u + \int \frac{1}{w^{2} + 3} d w)$

Step 9

Simplify.

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Step 9.1

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

$2 (\ln (| w |) + C) - (\int \frac{1}{u \cdot 2} d u + \int \frac{1}{w^{2} + 3} d w)$

Step 9.2

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

$2 (\ln (| w |) + C) - (\int \frac{1}{2 u} d u + \int \frac{1}{w^{2} + 3} d w)$

Step 10

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

$2 (\ln (| w |) + C) - (\frac{1}{2} \int \frac{1}{u} d u + \int \frac{1}{w^{2} + 3} d w)$

Step 11

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

$2 (\ln (| w |) + C) - (\frac{1}{2} (\ln (| u |) + C) + \int \frac{1}{w^{2} + 3} d w)$

Step 12

Reorder $w^{2}$ and $3$ .

$2 (\ln (| w |) + C) - (\frac{1}{2} (\ln (| u |) + C) + \int \frac{1}{3 + w^{2}} d w)$

Step 13

Rewrite $3$ as ${\sqrt{3}}^{2}$ .

$2 (\ln (| w |) + C) - (\frac{1}{2} (\ln (| u |) + C) + \int \frac{1}{{\sqrt{3}}^{2} + w^{2}} d w)$

Step 14

The integral of $\frac{1}{{\sqrt{3}}^{2} + w^{2}}$ with respect to $w$ is $\frac{1}{\sqrt{3}} \arctan (\frac{w}{\sqrt{3}}) + C$ .

$2 (\ln (| w |) + C) - (\frac{1}{2} (\ln (| u |) + C) + \frac{1}{\sqrt{3}} \arctan (\frac{w}{\sqrt{3}}) + C)$

Step 15

Simplify.

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Step 15.1

Combine $\frac{1}{\sqrt{3}}$ and $\arctan (\frac{w}{\sqrt{3}})$ .

$2 (\ln (| w |) + C) - (\frac{1}{2} (\ln (| u |) + C) + \frac{\arctan (\frac{w}{\sqrt{3}})}{\sqrt{3}} + C)$

Step 15.2

Simplify.

$2 \ln (| w |) - (\frac{\ln (| u |)}{2} + \frac{\arctan (\frac{w \sqrt{3}}{3}) \sqrt{3}}{3}) + C$

Step 16

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

$2 \ln (| w |) - (\frac{\ln (| w^{2} + 3 |)}{2} + \frac{\arctan (\frac{w \sqrt{3}}{3}) \sqrt{3}}{3}) + C$

Step 17

Simplify.

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Step 17.1

To write $\frac{\ln (| w^{2} + 3 |)}{2}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$2 \ln (| w |) - (\frac{\ln (| w^{2} + 3 |)}{2} \cdot \frac{3}{3} + \frac{\arctan (\frac{w \sqrt{3}}{3}) \sqrt{3}}{3}) + C$

Step 17.2

To write $\frac{\arctan (\frac{w \sqrt{3}}{3}) \sqrt{3}}{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$2 \ln (| w |) - (\frac{\ln (| w^{2} + 3 |)}{2} \cdot \frac{3}{3} + \frac{\arctan (\frac{w \sqrt{3}}{3}) \sqrt{3}}{3} \cdot \frac{2}{2}) + C$

Step 17.3

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Step 17.3.1

Multiply $\frac{\ln (| w^{2} + 3 |)}{2}$ by $\frac{3}{3}$ .

$2 \ln (| w |) - (\frac{\ln (| w^{2} + 3 |) \cdot 3}{2 \cdot 3} + \frac{\arctan (\frac{w \sqrt{3}}{3}) \sqrt{3}}{3} \cdot \frac{2}{2}) + C$

Step 17.3.2

Multiply $2$ by $3$ .

$2 \ln (| w |) - (\frac{\ln (| w^{2} + 3 |) \cdot 3}{6} + \frac{\arctan (\frac{w \sqrt{3}}{3}) \sqrt{3}}{3} \cdot \frac{2}{2}) + C$

Step 17.3.3

Multiply $\frac{\arctan (\frac{w \sqrt{3}}{3}) \sqrt{3}}{3}$ by $\frac{2}{2}$ .

$2 \ln (| w |) - (\frac{\ln (| w^{2} + 3 |) \cdot 3}{6} + \frac{\arctan (\frac{w \sqrt{3}}{3}) \sqrt{3} \cdot 2}{3 \cdot 2}) + C$

Step 17.3.4

Multiply $3$ by $2$ .

$2 \ln (| w |) - (\frac{\ln (| w^{2} + 3 |) \cdot 3}{6} + \frac{\arctan (\frac{w \sqrt{3}}{3}) \sqrt{3} \cdot 2}{6}) + C$

Step 17.4

Combine the numerators over the common denominator.

$2 \ln (| w |) - \frac{\ln (| w^{2} + 3 |) \cdot 3 + \arctan (\frac{w \sqrt{3}}{3}) \sqrt{3} \cdot 2}{6} + C$

Step 17.5

Simplify the numerator.

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Step 17.5.1

Move $3$ to the left of $\ln (| w^{2} + 3 |)$ .

$2 \ln (| w |) - \frac{3 \cdot \ln (| w^{2} + 3 |) + \arctan (\frac{w \sqrt{3}}{3}) \sqrt{3} \cdot 2}{6} + C$

Step 17.5.2

Move $2$ to the left of $\arctan (\frac{w \sqrt{3}}{3}) \sqrt{3}$ .

$2 \ln (| w |) - \frac{3 \ln (| w^{2} + 3 |) + 2 \arctan (\frac{w \sqrt{3}}{3}) \sqrt{3}}{6} + C$

Step 18

Reorder terms.

$2 \ln (| w |) - \frac{1}{6} (3 \ln (| w^{2} + 3 |) + 2 \arctan (\frac{\sqrt{3}}{3} w) \sqrt{3}) + C$