Calculus Examples

$\int_{0}^{1} \frac{2}{2 x^{2} + 3 x + 1} d x$

Step 1

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

$2 \int_{0}^{1} \frac{1}{2 x^{2} + 3 x + 1} d x$

Step 2

Write the fraction using partial fraction decomposition.

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Step 2.1

Decompose the fraction and multiply through by the common denominator.

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Step 2.1.1

Factor by grouping.

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Step 2.1.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 1 = 2$ and whose sum is $b = 3$ .

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Step 2.1.1.1.1

Factor $3$ out of $3 x$ .

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

Step 2.1.1.1.2

Rewrite $3$ as $1$ plus $2$

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

Step 2.1.1.1.3

Apply the distributive property.

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

Step 2.1.1.1.4

Multiply $x$ by $1$ .

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

Step 2.1.1.2

Factor out the greatest common factor from each group.

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Step 2.1.1.2.1

Group the first two terms and the last two terms.

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

Step 2.1.1.2.2

Factor out the greatest common factor (GCF) from each group.

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

Step 2.1.1.3

Factor the polynomial by factoring out the greatest common factor, $2 x + 1$ .

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

Step 2.1.2

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 $A$ .

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

Step 2.1.3

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

Step 2.1.4

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

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

Step 2.1.5

Cancel the common factor of $2 x + 1$ .

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Step 2.1.5.1

Cancel the common factor.

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

Step 2.1.5.2

Rewrite the expression.

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

Step 2.1.6

Cancel the common factor of $x + 1$ .

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Step 2.1.6.1

Cancel the common factor.

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

Step 2.1.6.2

Rewrite the expression.

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

Step 2.1.7

Simplify each term.

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Step 2.1.7.1

Cancel the common factor of $2 x + 1$ .

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Step 2.1.7.1.1

Cancel the common factor.

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

Step 2.1.7.1.2

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

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

Step 2.1.7.2

Apply the distributive property.

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

Step 2.1.7.3

Multiply $A$ by $1$ .

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

Step 2.1.7.4

Cancel the common factor of $x + 1$ .

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Step 2.1.7.4.1

Cancel the common factor.

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

Step 2.1.7.4.2

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

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

Step 2.1.7.5

Apply the distributive property.

$1 = A x + A + B (2 x) + B \cdot 1$

Step 2.1.7.6

Rewrite using the commutative property of multiplication.

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

Step 2.1.7.7

Multiply $B$ by $1$ .

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

Step 2.1.8

Simplify the expression.

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Step 2.1.8.1

Move $B$ .

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

Step 2.1.8.2

Move $A$ .

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

Step 2.2

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

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Step 2.2.1

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.

$0 = A + 2 B$

Step 2.2.2

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.

$1 = A + B$

Step 2.2.3

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

$0 = A + 2 B$

$1 = A + B$

$0 = A + 2 B$

$1 = A + B$

Step 2.3

Solve the system of equations.

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Step 2.3.1

Solve for $A$ in $0 = A + 2 B$ .

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Step 2.3.1.1

Rewrite the equation as $A + 2 B = 0$ .

$A + 2 B = 0$

$1 = A + B$

Step 2.3.1.2

Subtract $2 B$ from both sides of the equation.

$A = - 2 B$

$1 = A + B$

$A = - 2 B$

$1 = A + B$

Step 2.3.2

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

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Step 2.3.2.1

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

$1 = (- 2 B) + B$

$A = - 2 B$

Step 2.3.2.2

Simplify the right side.

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Step 2.3.2.2.1

Add $- 2 B$ and $B$ .

$1 = - B$

$A = - 2 B$

$1 = - B$

$A = - 2 B$

$1 = - B$

$A = - 2 B$

Step 2.3.3

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

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Step 2.3.3.1

Rewrite the equation as $- B = 1$ .

$- B = 1$

$A = - 2 B$

Step 2.3.3.2

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

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Step 2.3.3.2.1

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

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

$A = - 2 B$

Step 2.3.3.2.2

Simplify the left side.

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Step 2.3.3.2.2.1

Dividing two negative values results in a positive value.

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

$A = - 2 B$

Step 2.3.3.2.2.2

Divide $B$ by $1$ .

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

$A = - 2 B$

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

$A = - 2 B$

Step 2.3.3.2.3

Simplify the right side.

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Step 2.3.3.2.3.1

Divide $1$ by $- 1$ .

$B = - 1$

$A = - 2 B$

$B = - 1$

$A = - 2 B$

$B = - 1$

$A = - 2 B$

$B = - 1$

$A = - 2 B$

Step 2.3.4

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

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Step 2.3.4.1

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

$A = - 2 \cdot - 1$

$B = - 1$

Step 2.3.4.2

Simplify the right side.

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Step 2.3.4.2.1

Multiply $- 2$ by $- 1$ .

$A = 2$

$B = - 1$

$A = 2$

$B = - 1$

$A = 2$

$B = - 1$

Step 2.3.5

List all of the solutions.

$A = 2, B = - 1$

Step 2.4

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

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

Step 2.5

Move the negative in front of the fraction.

$2 \int_{0}^{1} \frac{2}{2 x + 1} - \frac{1}{x + 1} d x$

Step 3

Split the single integral into multiple integrals.

$2 (\int_{0}^{1} \frac{2}{2 x + 1} d x + \int_{0}^{1} - \frac{1}{x + 1} d x)$

Step 4

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

$2 (2 \int_{0}^{1} \frac{1}{2 x + 1} d x + \int_{0}^{1} - \frac{1}{x + 1} d x)$

Step 5

Let $u_{1} = 2 x + 1$ . Then $d u_{1} = 2 d x$ , so $\frac{1}{2} d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 5.1

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

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Step 5.1.1

Differentiate $2 x + 1$ .

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

Step 5.1.2

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

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

Step 5.1.3

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

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Step 5.1.3.1

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

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

Step 5.1.3.2

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

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

Step 5.1.3.3

Multiply $2$ by $1$ .

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

Step 5.1.4

Differentiate using the Constant Rule.

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Step 5.1.4.1

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

$2 + 0$

Step 5.1.4.2

Add $2$ and $0$ .

$2$

Step 5.2

Substitute the lower limit in for $x$ in $u_{1} = 2 x + 1$ .

$u_{lower} = 2 \cdot 0 + 1$

Step 5.3

Simplify.

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Step 5.3.1

Multiply $2$ by $0$ .

$u_{lower} = 0 + 1$

Step 5.3.2

Add $0$ and $1$ .

$u_{lower} = 1$

Step 5.4

Substitute the upper limit in for $x$ in $u_{1} = 2 x + 1$ .

$u_{upper} = 2 \cdot 1 + 1$

Step 5.5

Simplify.

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Step 5.5.1

Multiply $2$ by $1$ .

$u_{upper} = 2 + 1$

Step 5.5.2

Add $2$ and $1$ .

$u_{upper} = 3$

Step 5.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 1$

$u_{upper} = 3$

Step 5.7

Rewrite the problem using $u_{1}$ , $d u_{1}$ , and the new limits of integration.

$2 (2 \int_{1}^{3} \frac{1}{u_{1}} \cdot \frac{1}{2} d u_{1} + \int_{0}^{1} - \frac{1}{x + 1} d x)$

Step 6

Simplify.

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Step 6.1

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

$2 (2 \int_{1}^{3} \frac{1}{u_{1} \cdot 2} d u_{1} + \int_{0}^{1} - \frac{1}{x + 1} d x)$

Step 6.2

Move $2$ to the left of $u_{1}$ .

$2 (2 \int_{1}^{3} \frac{1}{2 u_{1}} d u_{1} + \int_{0}^{1} - \frac{1}{x + 1} d x)$

Step 7

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

$2 (2 (\frac{1}{2} \int_{1}^{3} \frac{1}{u_{1}} d u_{1}) + \int_{0}^{1} - \frac{1}{x + 1} d x)$

Step 8

Simplify.

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Step 8.1

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

$2 (\frac{2}{2} \int_{1}^{3} \frac{1}{u_{1}} d u_{1} + \int_{0}^{1} - \frac{1}{x + 1} d x)$

Step 8.2

Cancel the common factor of $2$ .

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Step 8.2.1

Cancel the common factor.

$2 (\frac{2}{2} \int_{1}^{3} \frac{1}{u_{1}} d u_{1} + \int_{0}^{1} - \frac{1}{x + 1} d x)$

Step 8.2.2

Rewrite the expression.

$2 (1 \int_{1}^{3} \frac{1}{u_{1}} d u_{1} + \int_{0}^{1} - \frac{1}{x + 1} d x)$

Step 8.3

Multiply $\int_{1}^{3} \frac{1}{u_{1}} d u_{1}$ by $1$ .

$2 (\int_{1}^{3} \frac{1}{u_{1}} d u_{1} + \int_{0}^{1} - \frac{1}{x + 1} d x)$

Step 9

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

$2 (\ln (| u_{1} |)]_{1}^{3} + \int_{0}^{1} - \frac{1}{x + 1} d x)$

Step 10

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

$2 (\ln (| u_{1} |)]_{1}^{3} - \int_{0}^{1} \frac{1}{x + 1} d x)$

Step 11

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

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Step 11.1

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

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Step 11.1.1

Differentiate $x + 1$ .

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

Step 11.1.2

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 11.1.3

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 11.1.4

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

$1 + 0$

Step 11.1.5

Add $1$ and $0$ .

$1$

Step 11.2

Substitute the lower limit in for $x$ in $u_{2} = x + 1$ .

$u_{lower} = 0 + 1$

Step 11.3

Add $0$ and $1$ .

$u_{lower} = 1$

Step 11.4

Substitute the upper limit in for $x$ in $u_{2} = x + 1$ .

$u_{upper} = 1 + 1$

Step 11.5

Add $1$ and $1$ .

$u_{upper} = 2$

Step 11.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 1$

$u_{upper} = 2$

Step 11.7

Rewrite the problem using $u_{2}$ , $d u_{2}$ , and the new limits of integration.

$2 (\ln (| u_{1} |)]_{1}^{3} - \int_{1}^{2} \frac{1}{u_{2}} d u_{2})$

Step 12

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

$2 (\ln (| u_{1} |)]_{1}^{3} - (\ln (| u_{2} |)]_{1}^{2}))$

Step 13

Substitute and simplify.

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Step 13.1

Evaluate $\ln (| u_{1} |)$ at $3$ and at $1$ .

$2 ((\ln (| 3 |)) - \ln (| 1 |) - (\ln (| u_{2} |)]_{1}^{2}))$

Step 13.2

Evaluate $\ln (| u_{2} |)$ at $2$ and at $1$ .

$2 ((\ln (| 3 |)) - \ln (| 1 |) - ((\ln (| 2 |)) - \ln (| 1 |)))$

Step 13.3

Remove parentheses.

$2 (\ln (| 3 |) - \ln (| 1 |) - (\ln (| 2 |) - \ln (| 1 |)))$

Step 14

Simplify.

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Step 14.1

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$2 (\ln (\frac{| 3 |}{| 1 |}) - (\ln (| 2 |) - \ln (| 1 |)))$

Step 14.2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$2 (\ln (\frac{| 3 |}{| 1 |}) - \ln (\frac{| 2 |}{| 1 |}))$

Step 14.3

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$2 \ln (\frac{\frac{| 3 |}{| 1 |}}{\frac{| 2 |}{| 1 |}})$

Step 14.4

Rewrite $\frac{\frac{| 3 |}{| 1 |}}{\frac{| 2 |}{| 1 |}}$ as a product.

$2 \ln (\frac{| 3 |}{| 1 |} \cdot \frac{1}{\frac{| 2 |}{| 1 |}})$

Step 14.5

Multiply by the reciprocal of the fraction to divide by $\frac{| 2 |}{| 1 |}$ .

$2 \ln (\frac{| 3 |}{| 1 |} (1 \frac{| 1 |}{| 2 |}))$

Step 14.6

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

$2 \ln (\frac{| 3 |}{| 1 |} \cdot \frac{| 1 |}{| 2 |})$

Step 14.7

Multiply $\frac{| 3 |}{| 1 |}$ by $\frac{| 1 |}{| 2 |}$ .

$2 \ln (\frac{| 3 | | 1 |}{| 1 | | 2 |})$

Step 14.8

To multiply absolute values, multiply the terms inside each absolute value.

$2 \ln (\frac{| 3 \cdot 1 |}{| 1 | | 2 |})$

Step 14.9

Multiply $3$ by $1$ .

$2 \ln (\frac{| 3 |}{| 1 | | 2 |})$

Step 14.10

To multiply absolute values, multiply the terms inside each absolute value.

$2 \ln (\frac{| 3 |}{| 1 \cdot 2 |})$

Step 14.11

Multiply $2$ by $1$ .

$2 \ln (\frac{| 3 |}{| 2 |})$

Step 15

Simplify.

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Step 15.1

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$2 \ln (\frac{3}{| 2 |})$

Step 15.2

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$2 \ln (\frac{3}{2})$

Step 16

The result can be shown in multiple forms.

Exact Form:

$2 \ln (\frac{3}{2})$

Decimal Form:

$0.81093021 \dots$

Step 17