Calculus Examples

$\int_{0}^{\frac{1}{\sqrt{3}}} 4 \frac{t^{2} - 1}{t^{4} - 1} d t$

Step 1

Combine $4$ and $\frac{t^{2} - 1}{t^{4} - 1}$ .

$\int_{0}^{\frac{1}{\sqrt{3}}} \frac{4 (t^{2} - 1)}{t^{4} - 1} d t$

Step 2

Since $4$ is constant with respect to $t$ , move $4$ out of the integral.

$4 \int_{0}^{\frac{1}{\sqrt{3}}} \frac{t^{2} - 1}{t^{4} - 1} d t$

Step 3

Write the fraction using partial fraction decomposition.

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Step 3.1

Decompose the fraction and multiply through by the common denominator.

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Step 3.1.1

Factor the fraction.

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Step 3.1.1.1

Rewrite $1$ as $1^{2}$ .

$\frac{t^{2} - 1^{2}}{t^{4} - 1}$

Step 3.1.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = t$ and $b = 1$ .

$\frac{(t + 1) (t - 1)}{t^{4} - 1}$

Step 3.1.1.3

Rewrite $t^{4}$ as ${(t^{2})}^{2}$ .

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

Step 3.1.1.4

Rewrite $1$ as $1^{2}$ .

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

Step 3.1.1.5

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = t^{2}$ and $b = 1$ .

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

Step 3.1.1.6

Simplify.

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Step 3.1.1.6.1

Rewrite $1$ as $1^{2}$ .

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

Step 3.1.1.6.2

Factor.

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Step 3.1.1.6.2.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = t$ and $b = 1$ .

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

Step 3.1.1.6.2.2

Remove unnecessary parentheses.

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

Step 3.1.1.7

Reduce the expression $\frac{(t + 1) (t - 1)}{(t^{2} + 1) (t + 1) (t - 1)}$ by cancelling the common factors.

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Step 3.1.1.7.1

Cancel the common factor.

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

Step 3.1.1.7.2

Rewrite the expression.

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

Step 3.1.2

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 is 2nd order, $2$ terms are required in the numerator. The number of terms required in the numerator is always equal to the order of the factor in the denominator.

$\frac{A t + B}{t^{2} + 1}$

Step 3.1.3

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 $C$ .

$\frac{A t + B}{t^{2} + 1} + \frac{C}{t - 1}$

Step 3.1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(t^{2} + 1) (t - 1)$ .

$\frac{(t - 1) (t^{2} + 1) (t - 1)}{(t^{2} + 1) (t - 1)} = \frac{(A t + B) (t^{2} + 1) (t - 1)}{t^{2} + 1} + \frac{(C) (t^{2} + 1) (t - 1)}{t - 1}$

Step 3.1.5

Cancel the common factor of $t - 1$ .

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Step 3.1.5.1

Cancel the common factor.

$\frac{(t - 1) (t^{2} + 1) (t - 1)}{(t^{2} + 1) (t - 1)} = \frac{(A t + B) (t^{2} + 1) (t - 1)}{t^{2} + 1} + \frac{(C) (t^{2} + 1) (t - 1)}{t - 1}$

Step 3.1.5.2

Rewrite the expression.

$\frac{(t^{2} + 1) (t - 1)}{t^{2} + 1} = \frac{(A t + B) (t^{2} + 1) (t - 1)}{t^{2} + 1} + \frac{(C) (t^{2} + 1) (t - 1)}{t - 1}$

Step 3.1.6

Cancel the common factor of $t^{2} + 1$ .

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Step 3.1.6.1

Cancel the common factor.

$\frac{(t^{2} + 1) (t - 1)}{t^{2} + 1} = \frac{(A t + B) (t^{2} + 1) (t - 1)}{t^{2} + 1} + \frac{(C) (t^{2} + 1) (t - 1)}{t - 1}$

Step 3.1.6.2

Divide $t - 1$ by $1$ .

$t - 1 = \frac{(A t + B) (t^{2} + 1) (t - 1)}{t^{2} + 1} + \frac{(C) (t^{2} + 1) (t - 1)}{t - 1}$

Step 3.1.7

Simplify each term.

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Step 3.1.7.1

Cancel the common factor of $t^{2} + 1$ .

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Step 3.1.7.1.1

Cancel the common factor.

$t - 1 = \frac{(A t + B) (t^{2} + 1) (t - 1)}{t^{2} + 1} + \frac{(C) (t^{2} + 1) (t - 1)}{t - 1}$

Step 3.1.7.1.2

Divide $(A t + B) (t - 1)$ by $1$ .

$t - 1 = (A t + B) (t - 1) + \frac{(C) (t^{2} + 1) (t - 1)}{t - 1}$

Step 3.1.7.2

Expand $(A t + B) (t - 1)$ using the FOIL Method.

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Step 3.1.7.2.1

Apply the distributive property.

$t - 1 = A t (t - 1) + B (t - 1) + \frac{(C) (t^{2} + 1) (t - 1)}{t - 1}$

Step 3.1.7.2.2

Apply the distributive property.

$t - 1 = A t \cdot t + A t \cdot - 1 + B (t - 1) + \frac{(C) (t^{2} + 1) (t - 1)}{t - 1}$

Step 3.1.7.2.3

Apply the distributive property.

$t - 1 = A t \cdot t + A t \cdot - 1 + B t + B \cdot - 1 + \frac{(C) (t^{2} + 1) (t - 1)}{t - 1}$

Step 3.1.7.3

Simplify each term.

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Step 3.1.7.3.1

Multiply $t$ by $t$ by adding the exponents.

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Step 3.1.7.3.1.1

Move $t$ .

$t - 1 = A (t \cdot t) + A t \cdot - 1 + B t + B \cdot - 1 + \frac{(C) (t^{2} + 1) (t - 1)}{t - 1}$

Step 3.1.7.3.1.2

Multiply $t$ by $t$ .

$t - 1 = A t^{2} + A t \cdot - 1 + B t + B \cdot - 1 + \frac{(C) (t^{2} + 1) (t - 1)}{t - 1}$

Step 3.1.7.3.2

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

$t - 1 = A t^{2} - 1 \cdot (A t) + B t + B \cdot - 1 + \frac{(C) (t^{2} + 1) (t - 1)}{t - 1}$

Step 3.1.7.3.3

Rewrite $- 1 (A t)$ as $- (A t)$ .

$t - 1 = A t^{2} - (A t) + B t + B \cdot - 1 + \frac{(C) (t^{2} + 1) (t - 1)}{t - 1}$

Step 3.1.7.3.4

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

$t - 1 = A t^{2} - A t + B t - 1 \cdot B + \frac{(C) (t^{2} + 1) (t - 1)}{t - 1}$

Step 3.1.7.3.5

Rewrite $- 1 B$ as $- B$ .

$t - 1 = A t^{2} - A t + B t - B + \frac{(C) (t^{2} + 1) (t - 1)}{t - 1}$

Step 3.1.7.4

Cancel the common factor of $t - 1$ .

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Step 3.1.7.4.1

Cancel the common factor.

$t - 1 = A t^{2} - A t + B t - B + \frac{(C) (t^{2} + 1) (t - 1)}{t - 1}$

Step 3.1.7.4.2

Divide $(C) (t^{2} + 1)$ by $1$ .

$t - 1 = A t^{2} - A t + B t - B + (C) (t^{2} + 1)$

Step 3.1.7.5

Apply the distributive property.

$t - 1 = A t^{2} - A t + B t - B + C t^{2} + C \cdot 1$

Step 3.1.7.6

Multiply $C$ by $1$ .

$t - 1 = A t^{2} - A t + B t - B + C t^{2} + C$

Step 3.1.8

Simplify the expression.

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Step 3.1.8.1

Move $A$ .

$t - 1 = A t^{2} - 1 t A + B t - B + C t^{2} + C$

Step 3.1.8.2

Reorder $B$ and $t$ .

$t - 1 = A t^{2} - 1 t A + B t - B + C t^{2} + C$

Step 3.1.8.3

Move $- B$ .

$t - 1 = A t^{2} - 1 t A + B t + C t^{2} - B + C$

Step 3.1.8.4

Move $B t$ .

$t - 1 = A t^{2} - 1 t A + C t^{2} + B t - B + C$

Step 3.1.8.5

Move $- 1 t A$ .

$t - 1 = A t^{2} + C t^{2} - 1 t A + B t - B + C$

Step 3.2

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

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Step 3.2.1

Create an equation for the partial fraction variables by equating the coefficients of $t^{2}$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = A + C$

Step 3.2.2

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.

$1 = - 1 A + B$

Step 3.2.3

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.

$- 1 = - 1 B + C$

Step 3.2.4

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

$0 = A + C$

$1 = - 1 A + B$

$- 1 = - 1 B + C$

$0 = A + C$

$1 = - 1 A + B$

$- 1 = - 1 B + C$

Step 3.3

Solve the system of equations.

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Step 3.3.1

Solve for $A$ in $0 = A + C$ .

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Step 3.3.1.1

Rewrite the equation as $A + C = 0$ .

$A + C = 0$

$1 = - 1 A + B$

$- 1 = - 1 B + C$

Step 3.3.1.2

Subtract $C$ from both sides of the equation.

$A = - C$

$1 = - 1 A + B$

$- 1 = - 1 B + C$

$A = - C$

$1 = - 1 A + B$

$- 1 = - 1 B + C$

Step 3.3.2

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

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Step 3.3.2.1

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

$1 = - 1 (- C) + B$

$A = - C$

$- 1 = - 1 B + C$

Step 3.3.2.2

Simplify the right side.

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Step 3.3.2.2.1

Multiply $- 1 (- C)$ .

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Step 3.3.2.2.1.1

Multiply $- 1$ by $- 1$ .

$1 = 1 C + B$

$A = - C$

$- 1 = - 1 B + C$

Step 3.3.2.2.1.2

Multiply $C$ by $1$ .

$1 = C + B$

$A = - C$

$- 1 = - 1 B + C$

$1 = C + B$

$A = - C$

$- 1 = - 1 B + C$

$1 = C + B$

$A = - C$

$- 1 = - 1 B + C$

$1 = C + B$

$A = - C$

$- 1 = - 1 B + C$

Step 3.3.3

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

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Step 3.3.3.1

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

$C + B = 1$

$A = - C$

$- 1 = - 1 B + C$

Step 3.3.3.2

Subtract $B$ from both sides of the equation.

$C = 1 - B$

$A = - C$

$- 1 = - 1 B + C$

$C = 1 - B$

$A = - C$

$- 1 = - 1 B + C$

Step 3.3.4

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

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Step 3.3.4.1

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

$A = - (1 - B)$

$C = 1 - B$

$- 1 = - 1 B + C$

Step 3.3.4.2

Simplify the right side.

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Step 3.3.4.2.1

Simplify $- (1 - B)$ .

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Step 3.3.4.2.1.1

Apply the distributive property.

$A = - 1 \cdot 1 + B$

$C = 1 - B$

$- 1 = - 1 B + C$

Step 3.3.4.2.1.2

Multiply $- 1$ by $1$ .

$A = - 1 + B$

$C = 1 - B$

$- 1 = - 1 B + C$

Step 3.3.4.2.1.3

Multiply $- - B$ .

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Step 3.3.4.2.1.3.1

Multiply $- 1$ by $- 1$ .

$A = - 1 + 1 B$

$C = 1 - B$

$- 1 = - 1 B + C$

Step 3.3.4.2.1.3.2

Multiply $B$ by $1$ .

$A = - 1 + B$

$C = 1 - B$

$- 1 = - 1 B + C$

$A = - 1 + B$

$C = 1 - B$

$- 1 = - 1 B + C$

$A = - 1 + B$

$C = 1 - B$

$- 1 = - 1 B + C$

$A = - 1 + B$

$C = 1 - B$

$- 1 = - 1 B + C$

Step 3.3.4.3

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

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

$A = - 1 + B$

$C = 1 - B$

Step 3.3.4.4

Simplify $- 1 = - 1 B + 1 - B$ .

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Step 3.3.4.4.1

Simplify the left side.

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Step 3.3.4.4.1.1

Remove parentheses.

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

$A = - 1 + B$

$C = 1 - B$

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

$A = - 1 + B$

$C = 1 - B$

Step 3.3.4.4.2

Simplify the right side.

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Step 3.3.4.4.2.1

Simplify $- 1 B + 1 - B$ .

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Step 3.3.4.4.2.1.1

Rewrite $- 1 B$ as $- B$ .

$- 1 = - B + 1 - B$

$A = - 1 + B$

$C = 1 - B$

Step 3.3.4.4.2.1.2

Subtract $B$ from $- B$ .

$- 1 = - 2 B + 1$

$A = - 1 + B$

$C = 1 - B$

$- 1 = - 2 B + 1$

$A = - 1 + B$

$C = 1 - B$

$- 1 = - 2 B + 1$

$A = - 1 + B$

$C = 1 - B$

$- 1 = - 2 B + 1$

$A = - 1 + B$

$C = 1 - B$

$- 1 = - 2 B + 1$

$A = - 1 + B$

$C = 1 - B$

Step 3.3.5

Solve for $B$ in $- 1 = - 2 B + 1$ .

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Step 3.3.5.1

Rewrite the equation as $- 2 B + 1 = - 1$ .

$- 2 B + 1 = - 1$

$A = - 1 + B$

$C = 1 - B$

Step 3.3.5.2

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

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Step 3.3.5.2.1

Subtract $1$ from both sides of the equation.

$- 2 B = - 1 - 1$

$A = - 1 + B$

$C = 1 - B$

Step 3.3.5.2.2

Subtract $1$ from $- 1$ .

$- 2 B = - 2$

$A = - 1 + B$

$C = 1 - B$

$- 2 B = - 2$

$A = - 1 + B$

$C = 1 - B$

Step 3.3.5.3

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

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Step 3.3.5.3.1

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

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

$A = - 1 + B$

$C = 1 - B$

Step 3.3.5.3.2

Simplify the left side.

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Step 3.3.5.3.2.1

Cancel the common factor of $- 2$ .

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Step 3.3.5.3.2.1.1

Cancel the common factor.

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

$A = - 1 + B$

$C = 1 - B$

Step 3.3.5.3.2.1.2

Divide $B$ by $1$ .

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

$A = - 1 + B$

$C = 1 - B$

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

$A = - 1 + B$

$C = 1 - B$

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

$A = - 1 + B$

$C = 1 - B$

Step 3.3.5.3.3

Simplify the right side.

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Step 3.3.5.3.3.1

Divide $- 2$ by $- 2$ .

$B = 1$

$A = - 1 + B$

$C = 1 - B$

$B = 1$

$A = - 1 + B$

$C = 1 - B$

$B = 1$

$A = - 1 + B$

$C = 1 - B$

$B = 1$

$A = - 1 + B$

$C = 1 - B$

Step 3.3.6

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

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Step 3.3.6.1

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

$A = - 1 + 1$

$B = 1$

$C = 1 - B$

Step 3.3.6.2

Simplify $A = - 1 + 1$ .

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Step 3.3.6.2.1

Simplify the left side.

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Step 3.3.6.2.1.1

Remove parentheses.

$A = - 1 + 1$

$B = 1$

$C = 1 - B$

$A = - 1 + 1$

$B = 1$

$C = 1 - B$

Step 3.3.6.2.2

Simplify the right side.

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Step 3.3.6.2.2.1

Add $- 1$ and $1$ .

$A = 0$

$B = 1$

$C = 1 - B$

$A = 0$

$B = 1$

$C = 1 - B$

$A = 0$

$B = 1$

$C = 1 - B$

Step 3.3.6.3

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

$C = 1 - (1)$

$A = 0$

$B = 1$

Step 3.3.6.4

Simplify the right side.

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Step 3.3.6.4.1

Simplify $1 - (1)$ .

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Step 3.3.6.4.1.1

Multiply $- 1$ by $1$ .

$C = 1 - 1$

$A = 0$

$B = 1$

Step 3.3.6.4.1.2

Subtract $1$ from $1$ .

$C = 0$

$A = 0$

$B = 1$

$C = 0$

$A = 0$

$B = 1$

$C = 0$

$A = 0$

$B = 1$

$C = 0$

$A = 0$

$B = 1$

Step 3.3.7

List all of the solutions.

$C = 0, A = 0, B = 1$

Step 3.4

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

$\frac{0 t + 1}{t^{2} + 1} + \frac{0}{t - 1}$

Step 3.5

Simplify.

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Step 3.5.1

Remove parentheses.

$\frac{(0) \cdot t + 1}{t^{2} + 1} + \frac{0}{t - 1}$

Step 3.5.2

Simplify the numerator.

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Step 3.5.2.1

Multiply $0$ by $t$ .

$\frac{0 + 1}{t^{2} + 1} + \frac{0}{t - 1}$

Step 3.5.2.2

Add $0$ and $1$ .

$\frac{1}{t^{2} + 1} + \frac{0}{t - 1}$

Step 3.5.3

Divide $0$ by $t - 1$ .

$\frac{1}{t^{2} + 1} + 0$

Step 3.5.4

Remove the zero from the expression.

$4 \int_{0}^{\frac{1}{\sqrt{3}}} \frac{1}{t^{2} + 1} d t$

Step 4

Simplify the expression.

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Step 4.1

Reorder $t^{2}$ and $1$ .

$4 \int_{0}^{\frac{1}{\sqrt{3}}} \frac{1}{1 + t^{2}} d t$

Step 4.2

Rewrite $1$ as $1^{2}$ .

$4 \int_{0}^{\frac{1}{\sqrt{3}}} \frac{1}{1^{2} + t^{2}} d t$

Step 5

The integral of $\frac{1}{1^{2} + t^{2}}$ with respect to $t$ is $\arctan (t)]_{0}^{\frac{1}{\sqrt{3}}}$ .

$4 (\arctan (t)]_{0}^{\frac{1}{\sqrt{3}}})$

Step 6

Evaluate $\arctan (t)$ at $\frac{1}{\sqrt{3}}$ and at $0$ .

$4 (\arctan (\frac{1}{\sqrt{3}}) - \arctan (0))$

Step 7

Simplify.

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Step 7.1

Simplify each term.

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Step 7.1.1

Evaluate $\arctan (\frac{1}{\sqrt{3}})$ .

$4 (\frac{π}{6} - \arctan (0))$

Step 7.1.2

The exact value of $\arctan (0)$ is $0$ .

$4 (\frac{π}{6} - 0)$

Step 7.1.3

Multiply $- 1$ by $0$ .

$4 (\frac{π}{6} + 0)$

Step 7.2

Add $\frac{π}{6}$ and $0$ .

$4 \frac{π}{6}$

Step 7.3

Cancel the common factor of $2$ .

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Step 7.3.1

Factor $2$ out of $4$ .

$2 (2) \frac{π}{6}$

Step 7.3.2

Factor $2$ out of $6$ .

$2 \cdot 2 \frac{π}{2 \cdot 3}$

Step 7.3.3

Cancel the common factor.

$2 \cdot 2 \frac{π}{2 \cdot 3}$

Step 7.3.4

Rewrite the expression.

$2 \frac{π}{3}$

Step 7.4

Combine $2$ and $\frac{π}{3}$ .

$\frac{2 π}{3}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$\frac{2 π}{3}$

Decimal Form:

$2.09439510 \dots$

Step 9