Calculus Examples

$\int_{0}^{2} \frac{2 t}{{(t - 3)}^{2}} d t$

Step 1

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

$2 \int_{0}^{2} \frac{t}{{(t - 3)}^{2}} d t$

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

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 $A$ .

$\frac{A}{{(t - 3)}^{2}}$

Step 2.1.2

$\frac{A}{{(t - 3)}^{2}} + \frac{B}{t - 3}$

Step 2.1.3

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

$\frac{t {(t - 3)}^{2}}{{(t - 3)}^{2}} = \frac{(A) {(t - 3)}^{2}}{{(t - 3)}^{2}} + \frac{(B) {(t - 3)}^{2}}{t - 3}$

Step 2.1.4

Cancel the common factor of ${(t - 3)}^{2}$ .

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Step 2.1.4.1

Cancel the common factor.

$\frac{t {(t - 3)}^{2}}{{(t - 3)}^{2}} = \frac{(A) {(t - 3)}^{2}}{{(t - 3)}^{2}} + \frac{(B) {(t - 3)}^{2}}{t - 3}$

Step 2.1.4.2

Divide $t$ by $1$ .

$t = \frac{(A) {(t - 3)}^{2}}{{(t - 3)}^{2}} + \frac{(B) {(t - 3)}^{2}}{t - 3}$

Step 2.1.5

Simplify each term.

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Step 2.1.5.1

Cancel the common factor of ${(t - 3)}^{2}$ .

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Step 2.1.5.1.1

Cancel the common factor.

$t = \frac{A {(t - 3)}^{2}}{{(t - 3)}^{2}} + \frac{(B) {(t - 3)}^{2}}{t - 3}$

Step 2.1.5.1.2

Divide $A$ by $1$ .

$t = A + \frac{(B) {(t - 3)}^{2}}{t - 3}$

Step 2.1.5.2

Cancel the common factor of ${(t - 3)}^{2}$ and $t - 3$ .

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Step 2.1.5.2.1

Factor $t - 3$ out of $(B) {(t - 3)}^{2}$ .

$t = A + \frac{(t - 3) (B (t - 3))}{t - 3}$

Step 2.1.5.2.2

Cancel the common factors.

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Step 2.1.5.2.2.1

Multiply by $1$ .

$t = A + \frac{(t - 3) (B (t - 3))}{(t - 3) \cdot 1}$

Step 2.1.5.2.2.2

Cancel the common factor.

$t = A + \frac{(t - 3) (B (t - 3))}{(t - 3) \cdot 1}$

Step 2.1.5.2.2.3

Rewrite the expression.

$t = A + \frac{B (t - 3)}{1}$

Step 2.1.5.2.2.4

Divide $B (t - 3)$ by $1$ .

$t = A + B (t - 3)$

Step 2.1.5.3

Apply the distributive property.

$t = A + B t + B \cdot - 3$

Step 2.1.5.4

Move $- 3$ to the left of $B$ .

$t = A + B t - 3 B$

Step 2.1.6

Reorder $A$ and $B t$ .

$t = B t + A - 3 B$

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 $t$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$1 = B$

Step 2.2.2

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing $t$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = A - 3 B$

Step 2.2.3

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

$1 = B$

$0 = A - 3 B$

$1 = B$

$0 = A - 3 B$

Step 2.3

Solve the system of equations.

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Step 2.3.1

Rewrite the equation as $B = 1$ .

$B = 1$

$0 = A - 3 B$

Step 2.3.2

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

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Step 2.3.2.1

Replace all occurrences of $B$ in $0 = A - 3 B$ with $1$ .

$0 = A - 3 \cdot 1$

$B = 1$

Step 2.3.2.2

Simplify the right side.

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Step 2.3.2.2.1

Multiply $- 3$ by $1$ .

$0 = A - 3$

$B = 1$

$0 = A - 3$

$B = 1$

$0 = A - 3$

$B = 1$

Step 2.3.3

Solve for $A$ in $0 = A - 3$ .

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Step 2.3.3.1

Rewrite the equation as $A - 3 = 0$ .

$A - 3 = 0$

$B = 1$

Step 2.3.3.2

Add $3$ to both sides of the equation.

$A = 3$

$B = 1$

$A = 3$

$B = 1$

Step 2.3.4

Solve the system of equations.

$A = 3 B = 1$

Step 2.3.5

List all of the solutions.

$A = 3, B = 1$

Step 2.4

Replace each of the partial fraction coefficients in $\frac{A}{{(t - 3)}^{2}} + \frac{B}{t - 3}$ with the values found for $A$ and $B$ .

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

Step 2.5

Remove the zero from the expression.

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

Step 3

Split the single integral into multiple integrals.

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

Step 4

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

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

Step 5

Let $u_{1} = t - 3$ . Then $d u_{1} = d t$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 5.1

Let $u_{1} = t - 3$ . Find $\frac{d u_{1}}{d t}$ .

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Step 5.1.1

Differentiate $t - 3$ .

$\frac{d}{d t} [t - 3]$

Step 5.1.2

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

$\frac{d}{d t} [t] + \frac{d}{d t} [- 3]$

Step 5.1.3

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

$1 + \frac{d}{d t} [- 3]$

Step 5.1.4

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

$1 + 0$

Step 5.1.5

Add $1$ and $0$ .

$1$

Step 5.2

Substitute the lower limit in for $t$ in $u_{1} = t - 3$ .

$u_{lower} = 0 - 3$

Step 5.3

Subtract $3$ from $0$ .

$u_{lower} = - 3$

Step 5.4

Substitute the upper limit in for $t$ in $u_{1} = t - 3$ .

$u_{upper} = 2 - 3$

Step 5.5

Subtract $3$ from $2$ .

$u_{upper} = - 1$

Step 5.6

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

$u_{lower} = - 3$

$u_{upper} = - 1$

Step 5.7

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

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

Step 6

Apply basic rules of exponents.

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Step 6.1

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

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

Step 6.2

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

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Step 6.2.1

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

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

Step 6.2.2

Multiply $2$ by $- 1$ .

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

Step 7

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

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

Step 8

Let $u_{2} = t - 3$ . Then $d u_{2} = d t$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 8.1

Let $u_{2} = t - 3$ . Find $\frac{d u_{2}}{d t}$ .

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Step 8.1.1

Differentiate $t - 3$ .

$\frac{d}{d t} [t - 3]$

Step 8.1.2

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

$\frac{d}{d t} [t] + \frac{d}{d t} [- 3]$

Step 8.1.3

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

$1 + \frac{d}{d t} [- 3]$

Step 8.1.4

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

$1 + 0$

Step 8.1.5

Add $1$ and $0$ .

$1$

Step 8.2

Substitute the lower limit in for $t$ in $u_{2} = t - 3$ .

$u_{lower} = 0 - 3$

Step 8.3

Subtract $3$ from $0$ .

$u_{lower} = - 3$

Step 8.4

Substitute the upper limit in for $t$ in $u_{2} = t - 3$ .

$u_{upper} = 2 - 3$

Step 8.5

Subtract $3$ from $2$ .

$u_{upper} = - 1$

Step 8.6

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

$u_{lower} = - 3$

$u_{upper} = - 1$

Step 8.7

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

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

Step 9

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

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

Step 10

Substitute and simplify.

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Step 10.1

Evaluate $- {u_{1}}^{- 1}$ at $- 1$ and at $- 3$ .

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

Step 10.2

Evaluate $\ln (| u_{2} |)$ at $- 1$ and at $- 3$ .

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

Step 10.3

Simplify.

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Step 10.3.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 10.3.2

Move the negative one from the denominator of $\frac{1}{- 1}$ .

$2 (3 (- (- 1 \cdot 1) + {(- 3)}^{- 1}) + (\ln (| - 1 |)) - \ln (| - 3 |))$

Step 10.3.3

Multiply $- 1$ by $1$ .

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

Step 10.3.4

Multiply $- 1$ by $- 1$ .

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

Step 10.3.5

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 10.3.6

Move the negative in front of the fraction.

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

Step 10.3.7

Write $1$ as a fraction with a common denominator.

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

Step 10.3.8

Combine the numerators over the common denominator.

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

Step 10.3.9

Subtract $1$ from $3$ .

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

Step 10.3.10

Combine $3$ and $\frac{2}{3}$ .

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

Step 10.3.11

Multiply $3$ by $2$ .

$2 (\frac{6}{3} + (\ln (| - 1 |)) - \ln (| - 3 |))$

Step 10.3.12

Cancel the common factor of $6$ and $3$ .

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Step 10.3.12.1

Factor $3$ out of $6$ .

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

Step 10.3.12.2

Cancel the common factors.

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Step 10.3.12.2.1

Factor $3$ out of $3$ .

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

Step 10.3.12.2.2

Cancel the common factor.

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

Step 10.3.12.2.3

Rewrite the expression.

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

Step 10.3.12.2.4

Divide $2$ by $1$ .

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

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

Step 11

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

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

Step 12

Simplify.

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Step 12.1

Simplify each term.

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Step 12.1.1

The absolute value is the distance between a number and zero. The distance between $- 1$ and $0$ is $1$ .

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

Step 12.1.2

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

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

Step 12.2

Apply the distributive property.

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

Step 12.3

Multiply $2$ by $2$ .

$4 + 2 \ln (\frac{1}{3})$

Step 13

The result can be shown in multiple forms.

Exact Form:

$4 + 2 \ln (\frac{1}{3})$

Decimal Form:

$1.80277542 \dots$

Step 14