Calculus Examples

$\int \frac{1}{(5 x - 3) (3 - 4 x)} d x$

Step 1

Write the fraction using partial fraction decomposition.

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Step 1.1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1.1

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 $A$ .

$\frac{A}{5 x - 3}$

Step 1.1.2

$\frac{A}{5 x - 3} + \frac{B}{3 - 4 x}$

Step 1.1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(5 x - 3) (3 - 4 x)$ .

$\frac{1 (5 x - 3) (3 - 4 x)}{(5 x - 3) (3 - 4 x)} = \frac{(A) (5 x - 3) (3 - 4 x)}{5 x - 3} + \frac{(B) (5 x - 3) (3 - 4 x)}{3 - 4 x}$

Step 1.1.4

Cancel the common factor of $5 x - 3$ .

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Step 1.1.4.1

Cancel the common factor.

$\frac{1 (5 x - 3) (3 - 4 x)}{(5 x - 3) (3 - 4 x)} = \frac{(A) (5 x - 3) (3 - 4 x)}{5 x - 3} + \frac{(B) (5 x - 3) (3 - 4 x)}{3 - 4 x}$

Step 1.1.4.2

Rewrite the expression.

$\frac{1 (3 - 4 x)}{3 - 4 x} = \frac{(A) (5 x - 3) (3 - 4 x)}{5 x - 3} + \frac{(B) (5 x - 3) (3 - 4 x)}{3 - 4 x}$

Step 1.1.5

Cancel the common factor of $3 - 4 x$ .

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Step 1.1.5.1

Cancel the common factor.

$\frac{1 (3 - 4 x)}{3 - 4 x} = \frac{(A) (5 x - 3) (3 - 4 x)}{5 x - 3} + \frac{(B) (5 x - 3) (3 - 4 x)}{3 - 4 x}$

Step 1.1.5.2

Rewrite the expression.

$1 = \frac{(A) (5 x - 3) (3 - 4 x)}{5 x - 3} + \frac{(B) (5 x - 3) (3 - 4 x)}{3 - 4 x}$

Step 1.1.6

Simplify each term.

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Step 1.1.6.1

Cancel the common factor of $5 x - 3$ .

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Step 1.1.6.1.1

Cancel the common factor.

$1 = \frac{A (5 x - 3) (3 - 4 x)}{5 x - 3} + \frac{(B) (5 x - 3) (3 - 4 x)}{3 - 4 x}$

Step 1.1.6.1.2

Divide $(A) (3 - 4 x)$ by $1$ .

$1 = (A) (3 - 4 x) + \frac{(B) (5 x - 3) (3 - 4 x)}{3 - 4 x}$

Step 1.1.6.2

Apply the distributive property.

$1 = A \cdot 3 + A (- 4 x) + \frac{(B) (5 x - 3) (3 - 4 x)}{3 - 4 x}$

Step 1.1.6.3

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

$1 = 3 \cdot A + A (- 4 x) + \frac{(B) (5 x - 3) (3 - 4 x)}{3 - 4 x}$

Step 1.1.6.4

Rewrite using the commutative property of multiplication.

$1 = 3 \cdot A - 4 A x + \frac{(B) (5 x - 3) (3 - 4 x)}{3 - 4 x}$

Step 1.1.6.5

Cancel the common factor of $3 - 4 x$ .

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Step 1.1.6.5.1

Cancel the common factor.

$1 = 3 A - 4 A x + \frac{(B) (5 x - 3) (3 - 4 x)}{3 - 4 x}$

Step 1.1.6.5.2

Divide $(B) (5 x - 3)$ by $1$ .

$1 = 3 A - 4 A x + (B) (5 x - 3)$

Step 1.1.6.6

Apply the distributive property.

$1 = 3 A - 4 A x + B (5 x) + B \cdot - 3$

Step 1.1.6.7

Rewrite using the commutative property of multiplication.

$1 = 3 A - 4 A x + 5 B x + B \cdot - 3$

Step 1.1.6.8

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

$1 = 3 A - 4 A x + 5 B x - 3 B$

Step 1.1.7

Reorder.

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Step 1.1.7.1

Move $A$ .

$1 = 3 A - 4 x A + 5 B x - 3 B$

Step 1.1.7.2

Move $B$ .

$1 = 3 A - 4 x A + 5 x B - 3 B$

Step 1.1.7.3

Move $3 A$ .

$1 = - 4 x A + 5 x B + 3 A - 3 B$

Step 1.2

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

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Step 1.2.1

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.

$0 = - 4 A + 5 B$

Step 1.2.2

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.

$1 = 3 A - 3 B$

Step 1.2.3

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

$0 = - 4 A + 5 B$

$1 = 3 A - 3 B$

$0 = - 4 A + 5 B$

$1 = 3 A - 3 B$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Solve for $A$ in $0 = - 4 A + 5 B$ .

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Step 1.3.1.1

Rewrite the equation as $- 4 A + 5 B = 0$ .

$- 4 A + 5 B = 0$

$1 = 3 A - 3 B$

Step 1.3.1.2

Subtract $5 B$ from both sides of the equation.

$- 4 A = - 5 B$

$1 = 3 A - 3 B$

Step 1.3.1.3

Divide each term in $- 4 A = - 5 B$ by $- 4$ and simplify.

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Step 1.3.1.3.1

Divide each term in $- 4 A = - 5 B$ by $- 4$ .

$\frac{- 4 A}{- 4} = \frac{- 5 B}{- 4}$

$1 = 3 A - 3 B$

Step 1.3.1.3.2

Simplify the left side.

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Step 1.3.1.3.2.1

Cancel the common factor of $- 4$ .

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Step 1.3.1.3.2.1.1

Cancel the common factor.

$\frac{- 4 A}{- 4} = \frac{- 5 B}{- 4}$

$1 = 3 A - 3 B$

Step 1.3.1.3.2.1.2

Divide $A$ by $1$ .

$A = \frac{- 5 B}{- 4}$

$1 = 3 A - 3 B$

$A = \frac{- 5 B}{- 4}$

$1 = 3 A - 3 B$

$A = \frac{- 5 B}{- 4}$

$1 = 3 A - 3 B$

Step 1.3.1.3.3

Simplify the right side.

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Step 1.3.1.3.3.1

Dividing two negative values results in a positive value.

$A = \frac{5 B}{4}$

$1 = 3 A - 3 B$

$A = \frac{5 B}{4}$

$1 = 3 A - 3 B$

$A = \frac{5 B}{4}$

$1 = 3 A - 3 B$

$A = \frac{5 B}{4}$

$1 = 3 A - 3 B$

Step 1.3.2

Replace all occurrences of $A$ with $\frac{5 B}{4}$ in each equation.

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Step 1.3.2.1

Replace all occurrences of $A$ in $1 = 3 A - 3 B$ with $\frac{5 B}{4}$ .

$1 = 3 (\frac{5 B}{4}) - 3 B$

$A = \frac{5 B}{4}$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Simplify $3 (\frac{5 B}{4}) - 3 B$ .

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Step 1.3.2.2.1.1

Multiply $3 (\frac{5 B}{4})$ .

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Step 1.3.2.2.1.1.1

Combine $3$ and $\frac{5 B}{4}$ .

$1 = \frac{3 (5 B)}{4} - 3 B$

$A = \frac{5 B}{4}$

Step 1.3.2.2.1.1.2

Multiply $5$ by $3$ .

$1 = \frac{15 B}{4} - 3 B$

$A = \frac{5 B}{4}$

$1 = \frac{15 B}{4} - 3 B$

$A = \frac{5 B}{4}$

Step 1.3.2.2.1.2

To write $- 3 B$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$1 = \frac{15 B}{4} - 3 B \cdot \frac{4}{4}$

$A = \frac{5 B}{4}$

Step 1.3.2.2.1.3

Simplify terms.

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Step 1.3.2.2.1.3.1

Combine $- 3 B$ and $\frac{4}{4}$ .

$1 = \frac{15 B}{4} + \frac{- 3 B \cdot 4}{4}$

$A = \frac{5 B}{4}$

Step 1.3.2.2.1.3.2

Combine the numerators over the common denominator.

$1 = \frac{15 B - 3 B \cdot 4}{4}$

$A = \frac{5 B}{4}$

$1 = \frac{15 B - 3 B \cdot 4}{4}$

$A = \frac{5 B}{4}$

Step 1.3.2.2.1.4

Simplify the numerator.

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Step 1.3.2.2.1.4.1

Factor $3 B$ out of $15 B - 3 B \cdot 4$ .

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Step 1.3.2.2.1.4.1.1

Factor $3 B$ out of $15 B$ .

$1 = \frac{3 B (5) - 3 B \cdot 4}{4}$

$A = \frac{5 B}{4}$

Step 1.3.2.2.1.4.1.2

Factor $3 B$ out of $- 3 B \cdot 4$ .

$1 = \frac{3 B (5) + 3 B (- 1 \cdot 4)}{4}$

$A = \frac{5 B}{4}$

Step 1.3.2.2.1.4.1.3

Factor $3 B$ out of $3 B (5) + 3 B (- 1 \cdot 4)$ .

$1 = \frac{3 B (5 - 1 \cdot 4)}{4}$

$A = \frac{5 B}{4}$

$1 = \frac{3 B (5 - 1 \cdot 4)}{4}$

$A = \frac{5 B}{4}$

Step 1.3.2.2.1.4.2

Multiply $- 1$ by $4$ .

$1 = \frac{3 B (5 - 4)}{4}$

$A = \frac{5 B}{4}$

Step 1.3.2.2.1.4.3

Subtract $4$ from $5$ .

$1 = \frac{3 B \cdot 1}{4}$

$A = \frac{5 B}{4}$

Step 1.3.2.2.1.4.4

Multiply $3$ by $1$ .

$1 = \frac{3 B}{4}$

$A = \frac{5 B}{4}$

$1 = \frac{3 B}{4}$

$A = \frac{5 B}{4}$

$1 = \frac{3 B}{4}$

$A = \frac{5 B}{4}$

$1 = \frac{3 B}{4}$

$A = \frac{5 B}{4}$

$1 = \frac{3 B}{4}$

$A = \frac{5 B}{4}$

Step 1.3.3

Solve for $B$ in $1 = \frac{3 B}{4}$ .

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Step 1.3.3.1

Rewrite the equation as $\frac{3 B}{4} = 1$ .

$\frac{3 B}{4} = 1$

$A = \frac{5 B}{4}$

Step 1.3.3.2

Multiply both sides of the equation by $\frac{4}{3}$ .

$\frac{4}{3} \cdot \frac{3 B}{4} = \frac{4}{3} \cdot 1$

$A = \frac{5 B}{4}$

Step 1.3.3.3

Simplify both sides of the equation.

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Step 1.3.3.3.1

Simplify the left side.

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Step 1.3.3.3.1.1

Simplify $\frac{4}{3} \cdot \frac{3 B}{4}$ .

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Step 1.3.3.3.1.1.1

Cancel the common factor of $4$ .

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Step 1.3.3.3.1.1.1.1

Cancel the common factor.

$\frac{4}{3} \cdot \frac{3 B}{4} = \frac{4}{3} \cdot 1$

$A = \frac{5 B}{4}$

Step 1.3.3.3.1.1.1.2

Rewrite the expression.

$\frac{1}{3} \cdot (3 B) = \frac{4}{3} \cdot 1$

$A = \frac{5 B}{4}$

$\frac{1}{3} \cdot (3 B) = \frac{4}{3} \cdot 1$

$A = \frac{5 B}{4}$

Step 1.3.3.3.1.1.2

Cancel the common factor of $3$ .

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Step 1.3.3.3.1.1.2.1

Factor $3$ out of $3 B$ .

$\frac{1}{3} \cdot (3 (B)) = \frac{4}{3} \cdot 1$

$A = \frac{5 B}{4}$

Step 1.3.3.3.1.1.2.2

Cancel the common factor.

$\frac{1}{3} \cdot (3 B) = \frac{4}{3} \cdot 1$

$A = \frac{5 B}{4}$

Step 1.3.3.3.1.1.2.3

Rewrite the expression.

$B = \frac{4}{3} \cdot 1$

$A = \frac{5 B}{4}$

$B = \frac{4}{3} \cdot 1$

$A = \frac{5 B}{4}$

$B = \frac{4}{3} \cdot 1$

$A = \frac{5 B}{4}$

$B = \frac{4}{3} \cdot 1$

$A = \frac{5 B}{4}$

Step 1.3.3.3.2

Simplify the right side.

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Step 1.3.3.3.2.1

Multiply $\frac{4}{3}$ by $1$ .

$B = \frac{4}{3}$

$A = \frac{5 B}{4}$

$B = \frac{4}{3}$

$A = \frac{5 B}{4}$

$B = \frac{4}{3}$

$A = \frac{5 B}{4}$

$B = \frac{4}{3}$

$A = \frac{5 B}{4}$

Step 1.3.4

Replace all occurrences of $B$ with $\frac{4}{3}$ in each equation.

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Step 1.3.4.1

Replace all occurrences of $B$ in $A = \frac{5 B}{4}$ with $\frac{4}{3}$ .

$A = \frac{5 (\frac{4}{3})}{4}$

$B = \frac{4}{3}$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

Simplify $\frac{5 (\frac{4}{3})}{4}$ .

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Step 1.3.4.2.1.1

Combine $5$ and $\frac{4}{3}$ .

$A = \frac{\frac{5 \cdot 4}{3}}{4}$

$B = \frac{4}{3}$

Step 1.3.4.2.1.2

Multiply $5$ by $4$ .

$A = \frac{\frac{20}{3}}{4}$

$B = \frac{4}{3}$

Step 1.3.4.2.1.3

Multiply the numerator by the reciprocal of the denominator.

$A = \frac{20}{3} \cdot \frac{1}{4}$

$B = \frac{4}{3}$

Step 1.3.4.2.1.4

Cancel the common factor of $4$ .

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Step 1.3.4.2.1.4.1

Factor $4$ out of $20$ .

$A = \frac{4 (5)}{3} \cdot \frac{1}{4}$

$B = \frac{4}{3}$

Step 1.3.4.2.1.4.2

Cancel the common factor.

$A = \frac{4 \cdot 5}{3} \cdot \frac{1}{4}$

$B = \frac{4}{3}$

Step 1.3.4.2.1.4.3

Rewrite the expression.

$A = \frac{5}{3}$

$B = \frac{4}{3}$

$A = \frac{5}{3}$

$B = \frac{4}{3}$

$A = \frac{5}{3}$

$B = \frac{4}{3}$

$A = \frac{5}{3}$

$B = \frac{4}{3}$

$A = \frac{5}{3}$

$B = \frac{4}{3}$

Step 1.3.5

List all of the solutions.

$A = \frac{5}{3}, B = \frac{4}{3}$

Step 1.4

Replace each of the partial fraction coefficients in $\frac{A}{5 x - 3} + \frac{B}{3 - 4 x}$ with the values found for $A$ and $B$ .

$\frac{\frac{5}{3}}{5 x - 3} + \frac{\frac{4}{3}}{3 - 4 x}$

Step 1.5

Simplify.

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Step 1.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{5}{3} \cdot \frac{1}{5 x - 3} + \frac{\frac{4}{3}}{3 - 4 x}$

Step 1.5.2

Multiply $\frac{5}{3}$ by $\frac{1}{5 x - 3}$ .

$\frac{5}{3 (5 x - 3)} + \frac{\frac{4}{3}}{3 - 4 x}$

Step 1.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{5}{3 (5 x - 3)} + \frac{4}{3} \cdot \frac{1}{3 - 4 x}$

Step 1.5.4

Multiply $\frac{4}{3}$ by $\frac{1}{3 - 4 x}$ .

$\int \frac{5}{3 (5 x - 3)} + \frac{4}{3 (3 - 4 x)} d x$

Step 2

Split the single integral into multiple integrals.

$\int \frac{5}{3 (5 x - 3)} d x + \int \frac{4}{3 (3 - 4 x)} d x$

Step 3

Since $\frac{5}{3}$ is constant with respect to $x$ , move $\frac{5}{3}$ out of the integral.

$\frac{5}{3} \int \frac{1}{5 x - 3} d x + \int \frac{4}{3 (3 - 4 x)} d x$

Step 4

Let $u_{1} = 5 x - 3$ . Then $d u_{1} = 5 d x$ , so $\frac{1}{5} d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 4.1

Let $u_{1} = 5 x - 3$ . Find $\frac{d u_{1}}{d x}$ .

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Step 4.1.1

Differentiate $5 x - 3$ .

$\frac{d}{d x} [5 x - 3]$

Step 4.1.2

By the Sum Rule, the derivative of $5 x - 3$ with respect to $x$ is $\frac{d}{d x} [5 x] + \frac{d}{d x} [- 3]$ .

$\frac{d}{d x} [5 x] + \frac{d}{d x} [- 3]$

Step 4.1.3

Evaluate $\frac{d}{d x} [5 x]$ .

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Step 4.1.3.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

$5 \frac{d}{d x} [x] + \frac{d}{d x} [- 3]$

Step 4.1.3.2

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

$5 \cdot 1 + \frac{d}{d x} [- 3]$

Step 4.1.3.3

Multiply $5$ by $1$ .

$5 + \frac{d}{d x} [- 3]$

Step 4.1.4

Differentiate using the Constant Rule.

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Step 4.1.4.1

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

$5 + 0$

Step 4.1.4.2

Add $5$ and $0$ .

$5$

Step 4.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\frac{5}{3} \int \frac{1}{u_{1}} \cdot \frac{1}{5} d u_{1} + \int \frac{4}{3 (3 - 4 x)} d x$

Step 5

Simplify.

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Step 5.1

Multiply $\frac{1}{u_{1}}$ by $\frac{1}{5}$ .

$\frac{5}{3} \int \frac{1}{u_{1} \cdot 5} d u_{1} + \int \frac{4}{3 (3 - 4 x)} d x$

Step 5.2

Move $5$ to the left of $u_{1}$ .

$\frac{5}{3} \int \frac{1}{5 u_{1}} d u_{1} + \int \frac{4}{3 (3 - 4 x)} d x$

Step 6

Since $\frac{1}{5}$ is constant with respect to $u_{1}$ , move $\frac{1}{5}$ out of the integral.

$\frac{5}{3} (\frac{1}{5} \int \frac{1}{u_{1}} d u_{1}) + \int \frac{4}{3 (3 - 4 x)} d x$

Step 7

Simplify.

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Step 7.1

Multiply $\frac{1}{5}$ by $\frac{5}{3}$ .

$\frac{5}{5 \cdot 3} \int \frac{1}{u_{1}} d u_{1} + \int \frac{4}{3 (3 - 4 x)} d x$

Step 7.2

Multiply $5$ by $3$ .

$\frac{5}{15} \int \frac{1}{u_{1}} d u_{1} + \int \frac{4}{3 (3 - 4 x)} d x$

Step 7.3

Cancel the common factor of $5$ and $15$ .

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Step 7.3.1

Factor $5$ out of $5$ .

$\frac{5 (1)}{15} \int \frac{1}{u_{1}} d u_{1} + \int \frac{4}{3 (3 - 4 x)} d x$

Step 7.3.2

Cancel the common factors.

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Step 7.3.2.1

Factor $5$ out of $15$ .

$\frac{5 \cdot 1}{5 \cdot 3} \int \frac{1}{u_{1}} d u_{1} + \int \frac{4}{3 (3 - 4 x)} d x$

Step 7.3.2.2

Cancel the common factor.

$\frac{5 \cdot 1}{5 \cdot 3} \int \frac{1}{u_{1}} d u_{1} + \int \frac{4}{3 (3 - 4 x)} d x$

Step 7.3.2.3

Rewrite the expression.

$\frac{1}{3} \int \frac{1}{u_{1}} d u_{1} + \int \frac{4}{3 (3 - 4 x)} d x$

Step 8

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

$\frac{1}{3} (\ln (| u_{1} |) + C) + \int \frac{4}{3 (3 - 4 x)} d x$

Step 9

Since $\frac{4}{3}$ is constant with respect to $x$ , move $\frac{4}{3}$ out of the integral.

$\frac{1}{3} (\ln (| u_{1} |) + C) + \frac{4}{3} \int \frac{1}{3 - 4 x} d x$

Step 10

Let $u_{2} = 3 - 4 x$ . Then $d u_{2} = - 4 d x$ , so $- \frac{1}{4} d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 10.1

Let $u_{2} = 3 - 4 x$ . Find $\frac{d u_{2}}{d x}$ .

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Step 10.1.1

Differentiate $3 - 4 x$ .

$\frac{d}{d x} [3 - 4 x]$

Step 10.1.2

Differentiate.

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Step 10.1.2.1

By the Sum Rule, the derivative of $3 - 4 x$ with respect to $x$ is $\frac{d}{d x} [3] + \frac{d}{d x} [- 4 x]$ .

$\frac{d}{d x} [3] + \frac{d}{d x} [- 4 x]$

Step 10.1.2.2

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

$0 + \frac{d}{d x} [- 4 x]$

Step 10.1.3

Evaluate $\frac{d}{d x} [- 4 x]$ .

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Step 10.1.3.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x$ with respect to $x$ is $- 4 \frac{d}{d x} [x]$ .

$0 - 4 \frac{d}{d x} [x]$

Step 10.1.3.2

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

$0 - 4 \cdot 1$

Step 10.1.3.3

Multiply $- 4$ by $1$ .

$0 - 4$

Step 10.1.4

Subtract $4$ from $0$ .

$- 4$

Step 10.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\frac{1}{3} (\ln (| u_{1} |) + C) + \frac{4}{3} \int \frac{1}{u_{2}} \cdot \frac{1}{- 4} d u_{2}$

Step 11

Simplify.

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Step 11.1

Move the negative in front of the fraction.

$\frac{1}{3} (\ln (| u_{1} |) + C) + \frac{4}{3} \int \frac{1}{u_{2}} (- \frac{1}{4}) d u_{2}$

Step 11.2

Multiply $\frac{1}{u_{2}}$ by $\frac{1}{4}$ .

$\frac{1}{3} (\ln (| u_{1} |) + C) + \frac{4}{3} \int - \frac{1}{u_{2} \cdot 4} d u_{2}$

Step 11.3

Move $4$ to the left of $u_{2}$ .

$\frac{1}{3} (\ln (| u_{1} |) + C) + \frac{4}{3} \int - \frac{1}{4 u_{2}} d u_{2}$

Step 12

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

$\frac{1}{3} (\ln (| u_{1} |) + C) + \frac{4}{3} (- \int \frac{1}{4 u_{2}} d u_{2})$

Step 13

Since $\frac{1}{4}$ is constant with respect to $u_{2}$ , move $\frac{1}{4}$ out of the integral.

$\frac{1}{3} (\ln (| u_{1} |) + C) + \frac{4}{3} (- (\frac{1}{4} \int \frac{1}{u_{2}} d u_{2}))$

Step 14

Simplify.

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Step 14.1

Multiply $\frac{4}{3}$ by $\frac{1}{4}$ .

$\frac{1}{3} (\ln (| u_{1} |) + C) - \frac{4}{3 \cdot 4} \int \frac{1}{u_{2}} d u_{2}$

Step 14.2

Multiply $3$ by $4$ .

$\frac{1}{3} (\ln (| u_{1} |) + C) - \frac{4}{12} \int \frac{1}{u_{2}} d u_{2}$

Step 14.3

Cancel the common factor of $4$ and $12$ .

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Step 14.3.1

Factor $4$ out of $4$ .

$\frac{1}{3} (\ln (| u_{1} |) + C) - \frac{4 (1)}{12} \int \frac{1}{u_{2}} d u_{2}$

Step 14.3.2

Cancel the common factors.

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Step 14.3.2.1

Factor $4$ out of $12$ .

$\frac{1}{3} (\ln (| u_{1} |) + C) - \frac{4 \cdot 1}{4 \cdot 3} \int \frac{1}{u_{2}} d u_{2}$

Step 14.3.2.2

Cancel the common factor.

$\frac{1}{3} (\ln (| u_{1} |) + C) - \frac{4 \cdot 1}{4 \cdot 3} \int \frac{1}{u_{2}} d u_{2}$

Step 14.3.2.3

Rewrite the expression.

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

Step 15

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

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

Step 16

Simplify.

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

Step 17

Substitute back in for each integration substitution variable.

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Step 17.1

Replace all occurrences of $u_{1}$ with $5 x - 3$ .

$\frac{1}{3} \ln (| 5 x - 3 |) - \frac{1}{3} \ln (| u_{2} |) + C$

Step 17.2

Replace all occurrences of $u_{2}$ with $3 - 4 x$ .

$\frac{1}{3} \ln (| 5 x - 3 |) - \frac{1}{3} \ln (| 3 - 4 x |) + C$