Calculus Examples

$\frac{4 x - 1}{{(2 x - 1)}^{2}}$

Step 1

Write $\frac{4 x - 1}{{(2 x - 1)}^{2}}$ as a function.

$f (x) = \frac{4 x - 1}{{(2 x - 1)}^{2}}$

Step 2

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Step 3

Set up the integral to solve.

$F (x) = \int \frac{4 x - 1}{{(2 x - 1)}^{2}} d x$

Step 4

Write the fraction using partial fraction decomposition.

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Step 4.1

Decompose the fraction and multiply through by the common denominator.

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Step 4.1.1

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 $A$ .

$\frac{A}{{(2 x - 1)}^{2}}$

Step 4.1.2

$\frac{A}{{(2 x - 1)}^{2}} + \frac{B}{2 x - 1}$

Step 4.1.3

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

$\frac{(4 x - 1) {(2 x - 1)}^{2}}{{(2 x - 1)}^{2}} = \frac{(A) {(2 x - 1)}^{2}}{{(2 x - 1)}^{2}} + \frac{(B) {(2 x - 1)}^{2}}{2 x - 1}$

Step 4.1.4

Cancel the common factor of ${(2 x - 1)}^{2}$ .

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Step 4.1.4.1

Cancel the common factor.

$\frac{(4 x - 1) {(2 x - 1)}^{2}}{{(2 x - 1)}^{2}} = \frac{(A) {(2 x - 1)}^{2}}{{(2 x - 1)}^{2}} + \frac{(B) {(2 x - 1)}^{2}}{2 x - 1}$

Step 4.1.4.2

Divide $4 x - 1$ by $1$ .

$4 x - 1 = \frac{(A) {(2 x - 1)}^{2}}{{(2 x - 1)}^{2}} + \frac{(B) {(2 x - 1)}^{2}}{2 x - 1}$

Step 4.1.5

Simplify each term.

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Step 4.1.5.1

Cancel the common factor of ${(2 x - 1)}^{2}$ .

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Step 4.1.5.1.1

Cancel the common factor.

$4 x - 1 = \frac{A {(2 x - 1)}^{2}}{{(2 x - 1)}^{2}} + \frac{(B) {(2 x - 1)}^{2}}{2 x - 1}$

Step 4.1.5.1.2

Divide $A$ by $1$ .

$4 x - 1 = A + \frac{(B) {(2 x - 1)}^{2}}{2 x - 1}$

Step 4.1.5.2

Cancel the common factor of ${(2 x - 1)}^{2}$ and $2 x - 1$ .

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Step 4.1.5.2.1

Factor $2 x - 1$ out of $(B) {(2 x - 1)}^{2}$ .

$4 x - 1 = A + \frac{(2 x - 1) (B (2 x - 1))}{2 x - 1}$

Step 4.1.5.2.2

Cancel the common factors.

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Step 4.1.5.2.2.1

Multiply by $1$ .

$4 x - 1 = A + \frac{(2 x - 1) (B (2 x - 1))}{(2 x - 1) \cdot 1}$

Step 4.1.5.2.2.2

Cancel the common factor.

$4 x - 1 = A + \frac{(2 x - 1) (B (2 x - 1))}{(2 x - 1) \cdot 1}$

Step 4.1.5.2.2.3

Rewrite the expression.

$4 x - 1 = A + \frac{B (2 x - 1)}{1}$

Step 4.1.5.2.2.4

Divide $B (2 x - 1)$ by $1$ .

$4 x - 1 = A + B (2 x - 1)$

Step 4.1.5.3

Apply the distributive property.

$4 x - 1 = A + B (2 x) + B \cdot - 1$

Step 4.1.5.4

Rewrite using the commutative property of multiplication.

$4 x - 1 = A + 2 B x + B \cdot - 1$

Step 4.1.5.5

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

$4 x - 1 = A + 2 B x - 1 \cdot B$

Step 4.1.5.6

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

$4 x - 1 = A + 2 B x - B$

Step 4.1.6

Simplify the expression.

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Step 4.1.6.1

Move $B$ .

$4 x - 1 = A + 2 x B - B$

Step 4.1.6.2

Reorder $A$ and $2 x B$ .

$4 x - 1 = 2 x B + A - B$

Step 4.2

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

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Step 4.2.1

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.

$4 = 2 B$

Step 4.2.2

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.

$- 1 = A - 1 B$

Step 4.2.3

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

$4 = 2 B$

$- 1 = A - 1 B$

$4 = 2 B$

$- 1 = A - 1 B$

Step 4.3

Solve the system of equations.

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Step 4.3.1

Solve for $B$ in $4 = 2 B$ .

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Step 4.3.1.1

Rewrite the equation as $2 B = 4$ .

$2 B = 4$

$- 1 = A - 1 B$

Step 4.3.1.2

Divide each term in $2 B = 4$ by $2$ and simplify.

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Step 4.3.1.2.1

Divide each term in $2 B = 4$ by $2$ .

$\frac{2 B}{2} = \frac{4}{2}$

$- 1 = A - 1 B$

Step 4.3.1.2.2

Simplify the left side.

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Step 4.3.1.2.2.1

Cancel the common factor of $2$ .

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Step 4.3.1.2.2.1.1

Cancel the common factor.

$\frac{2 B}{2} = \frac{4}{2}$

$- 1 = A - 1 B$

Step 4.3.1.2.2.1.2

Divide $B$ by $1$ .

$B = \frac{4}{2}$

$- 1 = A - 1 B$

$B = \frac{4}{2}$

$- 1 = A - 1 B$

$B = \frac{4}{2}$

$- 1 = A - 1 B$

Step 4.3.1.2.3

Simplify the right side.

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Step 4.3.1.2.3.1

Divide $4$ by $2$ .

$B = 2$

$- 1 = A - 1 B$

$B = 2$

$- 1 = A - 1 B$

$B = 2$

$- 1 = A - 1 B$

$B = 2$

$- 1 = A - 1 B$

Step 4.3.2

Replace all occurrences of $B$ with $2$ in each equation.

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Step 4.3.2.1

Replace all occurrences of $B$ in $- 1 = A - 1 B$ with $2$ .

$- 1 = A - 1 \cdot 2$

$B = 2$

Step 4.3.2.2

Simplify the right side.

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Step 4.3.2.2.1

Multiply $- 1$ by $2$ .

$- 1 = A - 2$

$B = 2$

$- 1 = A - 2$

$B = 2$

$- 1 = A - 2$

$B = 2$

Step 4.3.3

Solve for $A$ in $- 1 = A - 2$ .

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Step 4.3.3.1

Rewrite the equation as $A - 2 = - 1$ .

$A - 2 = - 1$

$B = 2$

Step 4.3.3.2

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

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Step 4.3.3.2.1

Add $2$ to both sides of the equation.

$A = - 1 + 2$

$B = 2$

Step 4.3.3.2.2

Add $- 1$ and $2$ .

$A = 1$

$B = 2$

$A = 1$

$B = 2$

$A = 1$

$B = 2$

Step 4.3.4

Solve the system of equations.

$A = 1 B = 2$

Step 4.3.5

List all of the solutions.

$A = 1, B = 2$

Step 4.4

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

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

Step 4.5

Remove the zero from the expression.

$\int \frac{1}{{(2 x - 1)}^{2}} + \frac{2}{2 x - 1} d x$

Step 5

Split the single integral into multiple integrals.

$\int \frac{1}{{(2 x - 1)}^{2}} d x + \int \frac{2}{2 x - 1} d x$

Step 6

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

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Step 6.1

Let $u_{1} = 2 x - 1$ . Find $\frac{d u_{1}}{d x}$ .

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Step 6.1.1

Differentiate $2 x - 1$ .

$\frac{d}{d x} [2 x - 1]$

Step 6.1.2

By the Sum Rule, the derivative of $2 x - 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$ .

$\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$

Step 6.1.3

Evaluate $\frac{d}{d x} [2 x]$ .

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Step 6.1.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$2 \frac{d}{d x} [x] + \frac{d}{d x} [- 1]$

Step 6.1.3.2

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

$2 \cdot 1 + \frac{d}{d x} [- 1]$

Step 6.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d x} [- 1]$

Step 6.1.4

Differentiate using the Constant Rule.

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Step 6.1.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$2 + 0$

Step 6.1.4.2

Add $2$ and $0$ .

$2$

Step 6.2

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

$\int \frac{1}{{u_{1}}^{2}} \cdot \frac{1}{2} d u_{1} + \int \frac{2}{2 x - 1} d x$

Step 7

Simplify.

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Step 7.1

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

$\int \frac{1}{{u_{1}}^{2} \cdot 2} d u_{1} + \int \frac{2}{2 x - 1} d x$

Step 7.2

Move $2$ to the left of ${u_{1}}^{2}$ .

$\int \frac{1}{2 {u_{1}}^{2}} d u_{1} + \int \frac{2}{2 x - 1} d x$

Step 8

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

$\frac{1}{2} \int \frac{1}{{u_{1}}^{2}} d u_{1} + \int \frac{2}{2 x - 1} d x$

Step 9

Apply basic rules of exponents.

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Step 9.1

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

$\frac{1}{2} \int {({u_{1}}^{2})}^{- 1} d u_{1} + \int \frac{2}{2 x - 1} d x$

Step 9.2

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

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Step 9.2.1

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

$\frac{1}{2} \int {u_{1}}^{2 \cdot - 1} d u_{1} + \int \frac{2}{2 x - 1} d x$

Step 9.2.2

Multiply $2$ by $- 1$ .

$\frac{1}{2} \int {u_{1}}^{- 2} d u_{1} + \int \frac{2}{2 x - 1} d x$

Step 10

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

$\frac{1}{2} (- {u_{1}}^{- 1} + C) + \int \frac{2}{2 x - 1} d x$

Step 11

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

$\frac{1}{2} (- {u_{1}}^{- 1} + C) + 2 \int \frac{1}{2 x - 1} d x$

Step 12

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

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Step 12.1

Let $u_{2} = 2 x - 1$ . Find $\frac{d u_{2}}{d x}$ .

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Step 12.1.1

Differentiate $2 x - 1$ .

$\frac{d}{d x} [2 x - 1]$

Step 12.1.2

By the Sum Rule, the derivative of $2 x - 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$ .

$\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$

Step 12.1.3

Evaluate $\frac{d}{d x} [2 x]$ .

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Step 12.1.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$2 \frac{d}{d x} [x] + \frac{d}{d x} [- 1]$

Step 12.1.3.2

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

$2 \cdot 1 + \frac{d}{d x} [- 1]$

Step 12.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d x} [- 1]$

Step 12.1.4

Differentiate using the Constant Rule.

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Step 12.1.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$2 + 0$

Step 12.1.4.2

Add $2$ and $0$ .

$2$

Step 12.2

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

$\frac{1}{2} (- {u_{1}}^{- 1} + C) + 2 \int \frac{1}{u_{2}} \cdot \frac{1}{2} d u_{2}$

Step 13

Simplify.

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Step 13.1

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

$\frac{1}{2} (- {u_{1}}^{- 1} + C) + 2 \int \frac{1}{u_{2} \cdot 2} d u_{2}$

Step 13.2

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

$\frac{1}{2} (- {u_{1}}^{- 1} + C) + 2 \int \frac{1}{2 u_{2}} d u_{2}$

Step 14

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

$\frac{1}{2} (- {u_{1}}^{- 1} + C) + 2 (\frac{1}{2} \int \frac{1}{u_{2}} d u_{2})$

Step 15

Simplify.

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Step 15.1

Combine $\frac{1}{2}$ and $2$ .

$\frac{1}{2} (- {u_{1}}^{- 1} + C) + \frac{2}{2} \int \frac{1}{u_{2}} d u_{2}$

Step 15.2

Cancel the common factor of $2$ .

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Step 15.2.1

Cancel the common factor.

$\frac{1}{2} (- {u_{1}}^{- 1} + C) + \frac{2}{2} \int \frac{1}{u_{2}} d u_{2}$

Step 15.2.2

Rewrite the expression.

$\frac{1}{2} (- {u_{1}}^{- 1} + C) + 1 \int \frac{1}{u_{2}} d u_{2}$

Step 15.3

Multiply $\int \frac{1}{u_{2}} d u_{2}$ by $1$ .

$\frac{1}{2} (- {u_{1}}^{- 1} + C) + \int \frac{1}{u_{2}} d u_{2}$

Step 16

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

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

Step 17

Simplify.

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Step 17.1

Simplify.

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

Step 17.2

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

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

Step 18

Substitute back in for each integration substitution variable.

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Step 18.1

Replace all occurrences of $u_{1}$ with $2 x - 1$ .

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

Step 18.2

Replace all occurrences of $u_{2}$ with $2 x - 1$ .

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

Step 19

The answer is the antiderivative of the function $f (x) = \frac{4 x - 1}{{(2 x - 1)}^{2}}$ .

$F (x) =$ $- \frac{1}{2 (2 x - 1)} + \ln (| 2 x - 1 |) + C$