Calculus Examples

$\int \frac{t^{3} + 5 t^{2} + 7 t + 1}{t^{2} + 3 t} d t$

Step 1

Divide $t^{3} + 5 t^{2} + 7 t + 1$ by $t^{2} + 3 t$ .

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Step 1.1

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

$t^{2}$

$3 t$

$0$

$t^{3}$

$5 t^{2}$

$7 t$

$1$

Step 1.2

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

$t$

$t^{2}$

$3 t$

$0$

$t^{3}$

$5 t^{2}$

$7 t$

$1$

Step 1.3

Multiply the new quotient term by the divisor.

$t$

$t^{2}$

$3 t$

$0$

$t^{3}$

$5 t^{2}$

$7 t$

$1$

$t^{3}$

$3 t^{2}$

$0$

Step 1.4

The expression needs to be subtracted from the dividend, so change all the signs in $t^{3} + 3 t^{2} + 0$

$t$

$t^{2}$

$3 t$

$0$

$t^{3}$

$5 t^{2}$

$7 t$

$1$

$t^{3}$

$3 t^{2}$

$0$

Step 1.5

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

$t$

$t^{2}$

$3 t$

$0$

$t^{3}$

$5 t^{2}$

$7 t$

$1$

$t^{3}$

$3 t^{2}$

$0$

$2 t^{2}$

$7 t$

Step 1.6

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

$t$

$t^{2}$

$3 t$

$0$

$t^{3}$

$5 t^{2}$

$7 t$

$1$

$t^{3}$

$3 t^{2}$

$0$

$2 t^{2}$

$7 t$

$1$

Step 1.7

Divide the highest order term in the dividend $2 t^{2}$ by the highest order term in divisor $t^{2}$ .

$t$

$2$

$t^{2}$

$3 t$

$0$

$t^{3}$

$5 t^{2}$

$7 t$

$1$

$t^{3}$

$3 t^{2}$

$0$

$2 t^{2}$

$7 t$

$1$

Step 1.8

Multiply the new quotient term by the divisor.

$t$

$2$

$t^{2}$

$3 t$

$0$

$t^{3}$

$5 t^{2}$

$7 t$

$1$

$t^{3}$

$3 t^{2}$

$0$

$2 t^{2}$

$7 t$

$1$

$2 t^{2}$

$6 t$

$0$

Step 1.9

The expression needs to be subtracted from the dividend, so change all the signs in $2 t^{2} + 6 t + 0$

$t$

$2$

$t^{2}$

$3 t$

$0$

$t^{3}$

$5 t^{2}$

$7 t$

$1$

$t^{3}$

$3 t^{2}$

$0$

$2 t^{2}$

$7 t$

$1$

$2 t^{2}$

$6 t$

$0$

Step 1.10

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

$t$

$2$

$t^{2}$

$3 t$

$0$

$t^{3}$

$5 t^{2}$

$7 t$

$1$

$t^{3}$

$3 t^{2}$

$0$

$2 t^{2}$

$7 t$

$1$

$2 t^{2}$

$6 t$

$0$

$t$

$1$

Step 1.11

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

$\int t + 2 + \frac{t + 1}{t^{2} + 3 t} d t$

Step 2

Split the single integral into multiple integrals.

$\int t d t + \int 2 d t + \int \frac{t + 1}{t^{2} + 3 t} d t$

Step 3

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

$\frac{1}{2} t^{2} + C + \int 2 d t + \int \frac{t + 1}{t^{2} + 3 t} d t$

Step 4

Apply the constant rule.

$\frac{1}{2} t^{2} + C + 2 t + C + \int \frac{t + 1}{t^{2} + 3 t} d t$

Step 5

Write the fraction using partial fraction decomposition.

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Step 5.1

Decompose the fraction and multiply through by the common denominator.

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Step 5.1.1

Factor $t$ out of $t^{2} + 3 t$ .

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Step 5.1.1.1

Factor $t$ out of $t^{2}$ .

$\frac{t + 1}{t \cdot t + 3 t}$

Step 5.1.1.2

Factor $t$ out of $3 t$ .

$\frac{t + 1}{t \cdot t + t \cdot 3}$

Step 5.1.1.3

Factor $t$ out of $t \cdot t + t \cdot 3$ .

$\frac{t + 1}{t (t + 3)}$

Step 5.1.2

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 $B$ .

$\frac{A}{t} + \frac{B}{t + 3}$

Step 5.1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $t (t + 3)$ .

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

Step 5.1.4

Cancel the common factor of $t$ .

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Step 5.1.4.1

Cancel the common factor.

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

Step 5.1.4.2

Rewrite the expression.

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

Step 5.1.5

Cancel the common factor of $t + 3$ .

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Step 5.1.5.1

Cancel the common factor.

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

Step 5.1.5.2

Divide $t + 1$ by $1$ .

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

Step 5.1.6

Simplify each term.

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Step 5.1.6.1

Cancel the common factor of $t$ .

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Step 5.1.6.1.1

Cancel the common factor.

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

Step 5.1.6.1.2

Divide $A (t + 3)$ by $1$ .

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

Step 5.1.6.2

Apply the distributive property.

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

Step 5.1.6.3

Move $3$ to the left of $A$ .

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

Step 5.1.6.4

Cancel the common factor of $t + 3$ .

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Step 5.1.6.4.1

Cancel the common factor.

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

Step 5.1.6.4.2

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

$t + 1 = A t + 3 A + (B) (t)$

$t + 1 = A t + 3 A + B t$

Step 5.1.7

Move $3 A$ .

$t + 1 = A t + B t + 3 A$

Step 5.2

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

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Step 5.2.1

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.

$1 = A + B$

Step 5.2.2

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.

$1 = 3 A$

Step 5.2.3

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

$1 = A + B$

$1 = 3 A$

$1 = A + B$

$1 = 3 A$

Step 5.3

Solve the system of equations.

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Step 5.3.1

Solve for $A$ in $1 = 3 A$ .

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Step 5.3.1.1

Rewrite the equation as $3 A = 1$ .

$3 A = 1$

$1 = A + B$

Step 5.3.1.2

Divide each term in $3 A = 1$ by $3$ and simplify.

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Step 5.3.1.2.1

Divide each term in $3 A = 1$ by $3$ .

$\frac{3 A}{3} = \frac{1}{3}$

$1 = A + B$

Step 5.3.1.2.2

Simplify the left side.

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Step 5.3.1.2.2.1

Cancel the common factor of $3$ .

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Step 5.3.1.2.2.1.1

Cancel the common factor.

$\frac{3 A}{3} = \frac{1}{3}$

$1 = A + B$

Step 5.3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{1}{3}$

$1 = A + B$

$A = \frac{1}{3}$

$1 = A + B$

$A = \frac{1}{3}$

$1 = A + B$

$A = \frac{1}{3}$

$1 = A + B$

$A = \frac{1}{3}$

$1 = A + B$

Step 5.3.2

Replace all occurrences of $A$ with $\frac{1}{3}$ in each equation.

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Step 5.3.2.1

Replace all occurrences of $A$ in $1 = A + B$ with $\frac{1}{3}$ .

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

$A = \frac{1}{3}$

Step 5.3.2.2

Simplify the right side.

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Step 5.3.2.2.1

Remove parentheses.

$1 = \frac{1}{3} + B$

$A = \frac{1}{3}$

$1 = \frac{1}{3} + B$

$A = \frac{1}{3}$

$1 = \frac{1}{3} + B$

$A = \frac{1}{3}$

Step 5.3.3

Solve for $B$ in $1 = \frac{1}{3} + B$ .

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Step 5.3.3.1

Rewrite the equation as $\frac{1}{3} + B = 1$ .

$\frac{1}{3} + B = 1$

$A = \frac{1}{3}$

Step 5.3.3.2

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

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Step 5.3.3.2.1

Subtract $\frac{1}{3}$ from both sides of the equation.

$B = 1 - \frac{1}{3}$

$A = \frac{1}{3}$

Step 5.3.3.2.2

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

$B = \frac{3}{3} - \frac{1}{3}$

$A = \frac{1}{3}$

Step 5.3.3.2.3

Combine the numerators over the common denominator.

$B = \frac{3 - 1}{3}$

$A = \frac{1}{3}$

Step 5.3.3.2.4

Subtract $1$ from $3$ .

$B = \frac{2}{3}$

$A = \frac{1}{3}$

$B = \frac{2}{3}$

$A = \frac{1}{3}$

$B = \frac{2}{3}$

$A = \frac{1}{3}$

Step 5.3.4

Solve the system of equations.

$B = \frac{2}{3} A = \frac{1}{3}$

Step 5.3.5

List all of the solutions.

$B = \frac{2}{3}, A = \frac{1}{3}$

Step 5.4

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

$\frac{\frac{1}{3}}{t} + \frac{\frac{2}{3}}{t + 3}$

Step 5.5

Simplify.

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Step 5.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{3} \cdot \frac{1}{t} + \frac{\frac{2}{3}}{t + 3}$

Step 5.5.2

Multiply $\frac{1}{3}$ by $\frac{1}{t}$ .

$\frac{1}{3 t} + \frac{\frac{2}{3}}{t + 3}$

Step 5.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{3 t} + \frac{2}{3} \cdot \frac{1}{t + 3}$

Step 5.5.4

Multiply $\frac{2}{3}$ by $\frac{1}{t + 3}$ .

$\frac{1}{2} t^{2} + C + 2 t + C + \int \frac{1}{3 t} + \frac{2}{3 (t + 3)} d t$

Step 6

Split the single integral into multiple integrals.

$\frac{1}{2} t^{2} + C + 2 t + C + \int \frac{1}{3 t} d t + \int \frac{2}{3 (t + 3)} d t$

Step 7

Since $\frac{1}{3}$ is constant with respect to $t$ , move $\frac{1}{3}$ out of the integral.

$\frac{1}{2} t^{2} + C + 2 t + C + \frac{1}{3} \int \frac{1}{t} d t + \int \frac{2}{3 (t + 3)} d t$

Step 8

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

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

Step 9

Since $\frac{2}{3}$ is constant with respect to $t$ , move $\frac{2}{3}$ out of the integral.

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

Step 10

Let $u = t + 3$ . Then $d u = d t$ . Rewrite using $u$ and $d$ $u$ .

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Step 10.1

Let $u = t + 3$ . Find $\frac{d u}{d t}$ .

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Step 10.1.1

Differentiate $t + 3$ .

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

Step 10.1.2

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 10.1.3

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 10.1.4

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

$1 + 0$

Step 10.1.5

Add $1$ and $0$ .

$1$

Step 10.2

Rewrite the problem using $u$ and $d u$ .

$\frac{1}{2} t^{2} + C + 2 t + C + \frac{1}{3} (\ln (| t |) + C) + \frac{2}{3} \int \frac{1}{u} d u$

Step 11

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

$\frac{1}{2} t^{2} + C + 2 t + C + \frac{1}{3} (\ln (| t |) + C) + \frac{2}{3} (\ln (| u |) + C)$

Step 12

Simplify.

$\frac{1}{2} t^{2} + 2 t + \frac{1}{3} \ln (| t |) + \frac{2}{3} \ln (| u |) + C$

Step 13

Replace all occurrences of $u$ with $t + 3$ .

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