Calculus Examples

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

Step 1

Subtract $3 x^{2}$ from $x^{2}$ .

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

Step 2

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

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

Step 3

Simplify by multiplying through.

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Step 3.1

Apply the distributive property.

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

Step 3.2

Apply the distributive property.

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

Step 4

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

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

Step 5

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 6

Add $2$ and $1$ .

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

Step 7

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

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

Step 8

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

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

Step 9

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 10

Add $1$ and $1$ .

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

Step 11

Divide $- 2 x^{3} + 2 x^{2} - 1 x$ by $x^{2} - 4 x + 4$ .

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Step 11.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x^{2}$

$4 x$

$4$

$2 x^{3}$

$2 x^{2}$

$1 x$

$0$

Step 11.2

Divide the highest order term in the dividend $- 2 x^{3}$ by the highest order term in divisor $x^{2}$ .

						-	$2 x$
	$x^{2}$	-	$4 x$	+	$4$	-	$2 x^{3}$	+	$2 x^{2}$	-	$1 x$	+	$0$

Step 11.3

Multiply the new quotient term by the divisor.

$2 x$

$x^{2}$

$4 x$

$4$

$2 x^{3}$

$2 x^{2}$

$1 x$

$0$

$2 x^{3}$

$8 x^{2}$

$8 x$

Step 11.4

The expression needs to be subtracted from the dividend, so change all the signs in $- 2 x^{3} + 8 x^{2} - 8 x$

$2 x$

$x^{2}$

$4 x$

$4$

$2 x^{3}$

$2 x^{2}$

$1 x$

$0$

$2 x^{3}$

$8 x^{2}$

$8 x$

Step 11.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$2 x$

$x^{2}$

$4 x$

$4$

$2 x^{3}$

$2 x^{2}$

$1 x$

$0$

$2 x^{3}$

$8 x^{2}$

$8 x$

$6 x^{2}$

$7 x$

Step 11.6

Pull the next terms from the original dividend down into the current dividend.

$2 x$

$x^{2}$

$4 x$

$4$

$2 x^{3}$

$2 x^{2}$

$1 x$

$0$

$2 x^{3}$

$8 x^{2}$

$8 x$

$6 x^{2}$

$7 x$

$0$

Step 11.7

Divide the highest order term in the dividend $- 6 x^{2}$ by the highest order term in divisor $x^{2}$ .

$2 x$

$6$

$x^{2}$

$4 x$

$4$

$2 x^{3}$

$2 x^{2}$

$1 x$

$0$

$2 x^{3}$

$8 x^{2}$

$8 x$

$6 x^{2}$

$7 x$

$0$

Step 11.8

Multiply the new quotient term by the divisor.

					-	$2 x$	-	$6$
$x^{2}$	-	$4 x$	+	$4$	-	$2 x^{3}$	+	$2 x^{2}$	-	$1 x$	+	$0$
					+	$2 x^{3}$	-	$8 x^{2}$	+	$8 x$
							-	$6 x^{2}$	+	$7 x$	+	$0$
							-	$6 x^{2}$	+	$24 x$	-	$24$

Step 11.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 6 x^{2} + 24 x - 24$

$2 x$

$6$

$x^{2}$

$4 x$

$4$

$2 x^{3}$

$2 x^{2}$

$1 x$

$0$

$2 x^{3}$

$8 x^{2}$

$8 x$

$6 x^{2}$

$7 x$

$0$

$6 x^{2}$

$24 x$

$24$

Step 11.10

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$2 x$

$6$

$x^{2}$

$4 x$

$4$

$2 x^{3}$

$2 x^{2}$

$1 x$

$0$

$2 x^{3}$

$8 x^{2}$

$8 x$

$6 x^{2}$

$7 x$

$0$

$6 x^{2}$

$24 x$

$24$

$17 x$

$24$

Step 11.11

The final answer is the quotient plus the remainder over the divisor.

$d \int - 2 x - 6 + \frac{- 17 x + 24}{x^{2} - 4 x + 4} d x$

Step 12

Split the single integral into multiple integrals.

$d (\int - 2 x d x + \int - 6 d x + \int \frac{- 17 x + 24}{x^{2} - 4 x + 4} d x)$

Step 13

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

$d (- 2 \int x d x + \int - 6 d x + \int \frac{- 17 x + 24}{x^{2} - 4 x + 4} d x)$

Step 14

By the Power Rule, the integral of $x$ with respect to $x$ is $\frac{1}{2} x^{2}$ .

$d (- 2 (\frac{1}{2} x^{2} + C) + \int - 6 d x + \int \frac{- 17 x + 24}{x^{2} - 4 x + 4} d x)$

Step 15

Apply the constant rule.

$d (- 2 (\frac{1}{2} x^{2} + C) - 6 x + C + \int \frac{- 17 x + 24}{x^{2} - 4 x + 4} d x)$

Step 16

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

$d (- 2 (\frac{x^{2}}{2} + C) - 6 x + C + \int \frac{- 17 x + 24}{x^{2} - 4 x + 4} d x)$

Step 17

Write the fraction using partial fraction decomposition.

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Step 17.1

Decompose the fraction and multiply through by the common denominator.

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Step 17.1.1

Factor using the perfect square rule.

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Step 17.1.1.1

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

$\frac{- 17 x + 24}{x^{2} - 4 x + 2^{2}}$

Step 17.1.1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot x \cdot 2$

Step 17.1.1.3

Rewrite the polynomial.

$\frac{- 17 x + 24}{x^{2} - 2 \cdot x \cdot 2 + 2^{2}}$

Step 17.1.1.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 2$ .

$\frac{- 17 x + 24}{{(x - 2)}^{2}}$

Step 17.1.2

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 in the denominator is linear, put a single variable in its place $A$ .

$\frac{A}{{(x - 2)}^{2}}$

Step 17.1.3

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

Step 17.1.4

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

$\frac{(- 17 x + 24) {(x - 2)}^{2}}{{(x - 2)}^{2}} = \frac{(A) {(x - 2)}^{2}}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2}}{x - 2}$

Step 17.1.5

Cancel the common factor of ${(x - 2)}^{2}$ .

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Step 17.1.5.1

Cancel the common factor.

$\frac{(- 17 x + 24) {(x - 2)}^{2}}{{(x - 2)}^{2}} = \frac{(A) {(x - 2)}^{2}}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2}}{x - 2}$

Step 17.1.5.2

Divide $- 17 x + 24$ by $1$ .

$- 17 x + 24 = \frac{(A) {(x - 2)}^{2}}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2}}{x - 2}$

Step 17.1.6

Simplify each term.

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Step 17.1.6.1

Cancel the common factor of ${(x - 2)}^{2}$ .

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Step 17.1.6.1.1

Cancel the common factor.

$- 17 x + 24 = \frac{A {(x - 2)}^{2}}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2}}{x - 2}$

Step 17.1.6.1.2

Divide $A$ by $1$ .

$- 17 x + 24 = A + \frac{(B) {(x - 2)}^{2}}{x - 2}$

Step 17.1.6.2

Cancel the common factor of ${(x - 2)}^{2}$ and $x - 2$ .

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Step 17.1.6.2.1

Factor $x - 2$ out of $(B) {(x - 2)}^{2}$ .

$- 17 x + 24 = A + \frac{(x - 2) (B (x - 2))}{x - 2}$

Step 17.1.6.2.2

Cancel the common factors.

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Step 17.1.6.2.2.1

Multiply by $1$ .

$- 17 x + 24 = A + \frac{(x - 2) (B (x - 2))}{(x - 2) \cdot 1}$

Step 17.1.6.2.2.2

Cancel the common factor.

$- 17 x + 24 = A + \frac{(x - 2) (B (x - 2))}{(x - 2) \cdot 1}$

Step 17.1.6.2.2.3

Rewrite the expression.

$- 17 x + 24 = A + \frac{B (x - 2)}{1}$

Step 17.1.6.2.2.4

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

$- 17 x + 24 = A + B (x - 2)$

Step 17.1.6.3

Apply the distributive property.

$- 17 x + 24 = A + B x + B \cdot - 2$

Step 17.1.6.4

Move $- 2$ to the left of $B$ .

$- 17 x + 24 = A + B x - 2 B$

Step 17.1.7

Reorder $A$ and $B x$ .

$- 17 x + 24 = B x + A - 2 B$

Step 17.2

Create equations for the partial fraction variables and use them to set up a system of equations.

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Step 17.2.1

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.

$- 17 = B$

Step 17.2.2

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.

$24 = A - 2 B$

Step 17.2.3

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

$- 17 = B$

$24 = A - 2 B$

$- 17 = B$

$24 = A - 2 B$

Step 17.3

Solve the system of equations.

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Step 17.3.1

Rewrite the equation as $B = - 17$ .

$B = - 17$

$24 = A - 2 B$

Step 17.3.2

Replace all occurrences of $B$ with $- 17$ in each equation.

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Step 17.3.2.1

Replace all occurrences of $B$ in $24 = A - 2 B$ with $- 17$ .

$24 = A - 2 \cdot - 17$

$B = - 17$

Step 17.3.2.2

Simplify the right side.

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Step 17.3.2.2.1

Multiply $- 2$ by $- 17$ .

$24 = A + 34$

$B = - 17$

$24 = A + 34$

$B = - 17$

$24 = A + 34$

$B = - 17$

Step 17.3.3

Solve for $A$ in $24 = A + 34$ .

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Step 17.3.3.1

Rewrite the equation as $A + 34 = 24$ .

$A + 34 = 24$

$B = - 17$

Step 17.3.3.2

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

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Step 17.3.3.2.1

Subtract $34$ from both sides of the equation.

$A = 24 - 34$

$B = - 17$

Step 17.3.3.2.2

Subtract $34$ from $24$ .

$A = - 10$

$B = - 17$

$A = - 10$

$B = - 17$

$A = - 10$

$B = - 17$

Step 17.3.4

Solve the system of equations.

$A = - 10 B = - 17$

Step 17.3.5

List all of the solutions.

$A = - 10, B = - 17$

Step 17.4

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

$\frac{- 10}{{(x - 2)}^{2}} + \frac{- 17}{x - 2}$

Step 17.5

Simplify.

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Step 17.5.1

Move the negative in front of the fraction.

$- \frac{10}{{(x - 2)}^{2}} + \frac{- 17}{x - 2}$

Step 17.5.2

Move the negative in front of the fraction.

$d (- 2 (\frac{x^{2}}{2} + C) - 6 x + C + \int - \frac{10}{{(x - 2)}^{2}} - \frac{17}{x - 2} d x)$

Step 18

Split the single integral into multiple integrals.

$d (- 2 (\frac{x^{2}}{2} + C) - 6 x + C + \int - \frac{10}{{(x - 2)}^{2}} d x + \int - \frac{17}{x - 2} d x)$

Step 19

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

$d (- 2 (\frac{x^{2}}{2} + C) - 6 x + C - \int \frac{10}{{(x - 2)}^{2}} d x + \int - \frac{17}{x - 2} d x)$

Step 20

Since $10$ is constant with respect to $x$ , move $10$ out of the integral.

$d (- 2 (\frac{x^{2}}{2} + C) - 6 x + C - (10 \int \frac{1}{{(x - 2)}^{2}} d x) + \int - \frac{17}{x - 2} d x)$

Step 21

Multiply $10$ by $- 1$ .

$d (- 2 (\frac{x^{2}}{2} + C) - 6 x + C - 10 \int \frac{1}{{(x - 2)}^{2}} d x + \int - \frac{17}{x - 2} d x)$

Step 22

Let $u_{1} = x - 2$ . Then $d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 22.1

Let $u_{1} = x - 2$ . Find $\frac{d u_{1}}{d x}$ .

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Step 22.1.1

Differentiate $x - 2$ .

$\frac{d}{d x} [x - 2]$

Step 22.1.2

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

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

Step 22.1.3

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

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

Step 22.1.4

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$1 + 0$

Step 22.1.5

Add $1$ and $0$ .

$1$

Step 22.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$d (- 2 (\frac{x^{2}}{2} + C) - 6 x + C - 10 \int \frac{1}{{u_{1}}^{2}} d u_{1} + \int - \frac{17}{x - 2} d x)$

Step 23

Apply basic rules of exponents.

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Step 23.1

Move ${u_{1}}^{2}$ out of the denominator by raising it to the $- 1$ power.

$d (- 2 (\frac{x^{2}}{2} + C) - 6 x + C - 10 \int {({u_{1}}^{2})}^{- 1} d u_{1} + \int - \frac{17}{x - 2} d x)$

Step 23.2

Multiply the exponents in ${({u_{1}}^{2})}^{- 1}$ .

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Step 23.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$d (- 2 (\frac{x^{2}}{2} + C) - 6 x + C - 10 \int {u_{1}}^{2 \cdot - 1} d u_{1} + \int - \frac{17}{x - 2} d x)$

Step 23.2.2

Multiply $2$ by $- 1$ .

$d (- 2 (\frac{x^{2}}{2} + C) - 6 x + C - 10 \int {u_{1}}^{- 2} d u_{1} + \int - \frac{17}{x - 2} d x)$

Step 24

By the Power Rule, the integral of ${u_{1}}^{- 2}$ with respect to $u_{1}$ is $- {u_{1}}^{- 1}$ .

$d (- 2 (\frac{x^{2}}{2} + C) - 6 x + C - 10 (- {u_{1}}^{- 1} + C) + \int - \frac{17}{x - 2} d x)$

Step 25

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

$d (- 2 (\frac{x^{2}}{2} + C) - 6 x + C - 10 (- {u_{1}}^{- 1} + C) - \int \frac{17}{x - 2} d x)$

Step 26

Since $17$ is constant with respect to $x$ , move $17$ out of the integral.

$d (- 2 (\frac{x^{2}}{2} + C) - 6 x + C - 10 (- {u_{1}}^{- 1} + C) - (17 \int \frac{1}{x - 2} d x))$

Step 27

Multiply $17$ by $- 1$ .

$d (- 2 (\frac{x^{2}}{2} + C) - 6 x + C - 10 (- {u_{1}}^{- 1} + C) - 17 \int \frac{1}{x - 2} d x)$

Step 28

Let $u_{2} = x - 2$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 28.1

Let $u_{2} = x - 2$ . Find $\frac{d u_{2}}{d x}$ .

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Step 28.1.1

Differentiate $x - 2$ .

$\frac{d}{d x} [x - 2]$

Step 28.1.2

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

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

Step 28.1.3

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

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

Step 28.1.4

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$1 + 0$

Step 28.1.5

Add $1$ and $0$ .

$1$

Step 28.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$d (- 2 (\frac{x^{2}}{2} + C) - 6 x + C - 10 (- {u_{1}}^{- 1} + C) - 17 \int \frac{1}{u_{2}} d u_{2})$

Step 29

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$d (- 2 (\frac{x^{2}}{2} + C) - 6 x + C - 10 (- {u_{1}}^{- 1} + C) - 17 (\ln (| u_{2} |) + C))$

Step 30

Simplify.

$d (- x^{2} - 6 x + \frac{10}{u_{1}} - 17 \ln (| u_{2} |)) + C$

Step 31

Substitute back in for each integration substitution variable.

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Step 31.1

Replace all occurrences of $u_{1}$ with $x - 2$ .

$d (- x^{2} - 6 x + \frac{10}{x - 2} - 17 \ln (| u_{2} |)) + C$

Step 31.2

Replace all occurrences of $u_{2}$ with $x - 2$ .

$d (- x^{2} - 6 x + \frac{10}{x - 2} - 17 \ln (| x - 2 |)) + C$