Calculus Examples

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

Step 1

Write $\frac{x + 1}{x^{2} - 3 x + 2}$ as a function.

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

Step 2

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Step 3

Set up the integral to solve.

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

Step 4

Write the fraction using partial fraction decomposition.

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Step 4.1

Decompose the fraction and multiply through by the common denominator.

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Step 4.1.1

Factor $x^{2} - 3 x + 2$ using the AC method.

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Step 4.1.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $2$ and whose sum is $- 3$ .

$- 2, - 1$

Step 4.1.1.2

Write the factored form using these integers.

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

Step 4.1.2

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 $A$ .

$\frac{A}{x - 2}$

Step 4.1.3

$\frac{A}{x - 2} + \frac{B}{x - 1}$

Step 4.1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(x - 2) (x - 1)$ .

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

Step 4.1.5

Cancel the common factor of $x - 2$ .

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Step 4.1.5.1

Cancel the common factor.

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

Step 4.1.5.2

Rewrite the expression.

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

Step 4.1.6

Cancel the common factor of $x - 1$ .

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Step 4.1.6.1

Cancel the common factor.

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

Step 4.1.6.2

Divide $x + 1$ by $1$ .

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

Step 4.1.7

Simplify each term.

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Step 4.1.7.1

Cancel the common factor of $x - 2$ .

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Step 4.1.7.1.1

Cancel the common factor.

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

Step 4.1.7.1.2

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

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

Step 4.1.7.2

Apply the distributive property.

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

Step 4.1.7.3

Move $- 1$ to the left of $A$ .

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

Step 4.1.7.4

Rewrite $- 1 A$ as $- A$ .

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

Step 4.1.7.5

Cancel the common factor of $x - 1$ .

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Step 4.1.7.5.1

Cancel the common factor.

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

Step 4.1.7.5.2

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

$x + 1 = A x - A + (B) (x - 2)$

Step 4.1.7.6

Apply the distributive property.

$x + 1 = A x - A + B x + B \cdot - 2$

Step 4.1.7.7

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

$x + 1 = A x - A + B x - 2 B$

Step 4.1.8

Move $- A$ .

$x + 1 = A x + B x - A - 2 B$

Step 4.2

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

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Step 4.2.1

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.

$1 = A + B$

Step 4.2.2

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.

$1 = - 1 A - 2 B$

Step 4.2.3

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

$1 = A + B$

$1 = - 1 A - 2 B$

$1 = A + B$

$1 = - 1 A - 2 B$

Step 4.3

Solve the system of equations.

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Step 4.3.1

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

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Step 4.3.1.1

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

$A + B = 1$

$1 = - 1 A - 2 B$

Step 4.3.1.2

Subtract $B$ from both sides of the equation.

$A = 1 - B$

$1 = - 1 A - 2 B$

$A = 1 - B$

$1 = - 1 A - 2 B$

Step 4.3.2

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

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Step 4.3.2.1

Replace all occurrences of $A$ in $1 = - 1 A - 2 B$ with $1 - B$ .

$1 = - 1 (1 - B) - 2 B$

$A = 1 - B$

Step 4.3.2.2

Simplify the right side.

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Step 4.3.2.2.1

Simplify $- 1 (1 - B) - 2 B$ .

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Step 4.3.2.2.1.1

Simplify each term.

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Step 4.3.2.2.1.1.1

Apply the distributive property.

$1 = - 1 \cdot 1 - 1 (- B) - 2 B$

$A = 1 - B$

Step 4.3.2.2.1.1.2

Multiply $- 1$ by $1$ .

$1 = - 1 - 1 (- B) - 2 B$

$A = 1 - B$

Step 4.3.2.2.1.1.3

Multiply $- 1 (- B)$ .

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Step 4.3.2.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

$1 = - 1 + 1 B - 2 B$

$A = 1 - B$

Step 4.3.2.2.1.1.3.2

Multiply $B$ by $1$ .

$1 = - 1 + B - 2 B$

$A = 1 - B$

$1 = - 1 + B - 2 B$

$A = 1 - B$

$1 = - 1 + B - 2 B$

$A = 1 - B$

Step 4.3.2.2.1.2

Subtract $2 B$ from $B$ .

$1 = - 1 - B$

$A = 1 - B$

$1 = - 1 - B$

$A = 1 - B$

$1 = - 1 - B$

$A = 1 - B$

$1 = - 1 - B$

$A = 1 - B$

Step 4.3.3

Solve for $B$ in $1 = - 1 - B$ .

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Step 4.3.3.1

Rewrite the equation as $- 1 - B = 1$ .

$- 1 - B = 1$

$A = 1 - B$

Step 4.3.3.2

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

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Step 4.3.3.2.1

Add $1$ to both sides of the equation.

$- B = 1 + 1$

$A = 1 - B$

Step 4.3.3.2.2

Add $1$ and $1$ .

$- B = 2$

$A = 1 - B$

$- B = 2$

$A = 1 - B$

Step 4.3.3.3

Divide each term in $- B = 2$ by $- 1$ and simplify.

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Step 4.3.3.3.1

Divide each term in $- B = 2$ by $- 1$ .

$\frac{- B}{- 1} = \frac{2}{- 1}$

$A = 1 - B$

Step 4.3.3.3.2

Simplify the left side.

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Step 4.3.3.3.2.1

Dividing two negative values results in a positive value.

$\frac{B}{1} = \frac{2}{- 1}$

$A = 1 - B$

Step 4.3.3.3.2.2

Divide $B$ by $1$ .

$B = \frac{2}{- 1}$

$A = 1 - B$

$B = \frac{2}{- 1}$

$A = 1 - B$

Step 4.3.3.3.3

Simplify the right side.

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Step 4.3.3.3.3.1

Divide $2$ by $- 1$ .

$B = - 2$

$A = 1 - B$

$B = - 2$

$A = 1 - B$

$B = - 2$

$A = 1 - B$

$B = - 2$

$A = 1 - B$

Step 4.3.4

Replace all occurrences of $B$ with $- 2$ in each equation.

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Step 4.3.4.1

Replace all occurrences of $B$ in $A = 1 - B$ with $- 2$ .

$A = 1 - (- 2)$

$B = - 2$

Step 4.3.4.2

Simplify the right side.

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Step 4.3.4.2.1

Simplify $1 - (- 2)$ .

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Step 4.3.4.2.1.1

Multiply $- 1$ by $- 2$ .

$A = 1 + 2$

$B = - 2$

Step 4.3.4.2.1.2

Add $1$ and $2$ .

$A = 3$

$B = - 2$

$A = 3$

$B = - 2$

$A = 3$

$B = - 2$

$A = 3$

$B = - 2$

Step 4.3.5

List all of the solutions.

$A = 3, B = - 2$

Step 4.4

Replace each of the partial fraction coefficients in $\frac{A}{x - 2} + \frac{B}{x - 1}$ with the values found for $A$ and $B$ .

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

Step 4.5

Move the negative in front of the fraction.

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

Step 5

Split the single integral into multiple integrals.

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

Step 6

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

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

Step 7

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

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Step 7.1

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

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Step 7.1.1

Differentiate $x - 2$ .

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

Step 7.1.2

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 7.1.3

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 7.1.4

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

$1 + 0$

Step 7.1.5

Add $1$ and $0$ .

$1$

Step 7.2

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

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

Step 8

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

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

Step 9

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

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

Step 10

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

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

Step 11

Multiply $2$ by $- 1$ .

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

Step 12

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

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Step 12.1

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

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Step 12.1.1

Differentiate $x - 1$ .

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

Step 12.1.2

By the Sum Rule, the derivative of $x - 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 1]$ .

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

Step 12.1.3

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

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

Step 12.1.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$1 + 0$

Step 12.1.5

Add $1$ and $0$ .

$1$

Step 12.2

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

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

Step 13

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

$3 (\ln (| u_{1} |) + C) - 2 (\ln (| u_{2} |) + C)$

Step 14

Simplify.

$3 \ln (| u_{1} |) - 2 \ln (| u_{2} |) + C$

Step 15

Substitute back in for each integration substitution variable.

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Step 15.1

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

$3 \ln (| x - 2 |) - 2 \ln (| u_{2} |) + C$

Step 15.2

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

$3 \ln (| x - 2 |) - 2 \ln (| x - 1 |) + C$

Step 16

The answer is the antiderivative of the function $f (x) = \frac{x + 1}{x^{2} - 3 x + 2}$ .

$F (x) =$ $3 \ln (| x - 2 |) - 2 \ln (| x - 1 |) + C$