Calculus Examples

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

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

Step 1

Since

2

$2$ is constant with respect to

t

$t$ , move

2

$2$ out of the integral.

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

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

Step 2

Write the fraction using partial fraction decomposition.

Tap for more steps...

Step 2.1

Decompose the fraction and multiply through by the common denominator.

Tap for more steps...

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

$A$ .

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

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

Step 2.1.2

B

$B$ .

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

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

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}

${(t - 3)}^{2}$ .

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

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

Step 2.1.4

Cancel the common factor of

{(t - 3)}^{2}

${(t - 3)}^{2}$ .

Tap for more steps...

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.

Tap for more steps...

Step 2.1.5.1

Cancel the common factor of

${(t - 3)}^{2}$ .

Tap for more steps...

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

$t - 3$ .

Tap for more steps...

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.

Tap for more steps...

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.

Tap for more steps...

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.

Tap for more steps...

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.

Tap for more steps...

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.

Tap for more steps...

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

$0 = A - 3$ .

Tap for more steps...

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

$d$

$u_{1}$ .

Tap for more steps...

Step 5.1

Let

$u_{1} = t - 3$ . Find

$\frac{d u_{1}}{d t}$ .

Tap for more steps...

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.

Tap for more steps...

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

${({u_{1}}^{2})}^{- 1}$ .

Tap for more steps...

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

$d$

$u_{2}$ .

Tap for more steps...

Step 8.1

Let

$u_{2} = t - 3$ . Find

$\frac{d u_{2}}{d t}$ .

Tap for more steps...

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.

Tap for more steps...

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.

Tap for more steps...

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

$3$ .

Tap for more steps...

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.

Tap for more steps...

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.

Tap for more steps...

Step 12.1

Simplify each term.

Tap for more steps...

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