Calculus Examples

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

Step 1

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

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

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.

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

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 $A$ .

$\frac{A}{1 + x}$

Step 4.1.2

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

Step 4.1.3

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

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

Step 4.1.4

Cancel the common factor of $1 + x$ .

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Step 4.1.4.1

Cancel the common factor.

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

Step 4.1.4.2

Rewrite the expression.

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

Step 4.1.5

Cancel the common factor of $1 - 2 x$ .

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Step 4.1.5.1

Cancel the common factor.

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

Step 4.1.5.2

Rewrite the expression.

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

Step 4.1.6

Simplify each term.

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Step 4.1.6.1

Cancel the common factor of $1 + x$ .

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Step 4.1.6.1.1

Cancel the common factor.

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

Step 4.1.6.1.2

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

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

Step 4.1.6.2

Apply the distributive property.

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

Step 4.1.6.3

Multiply $A$ by $1$ .

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

Step 4.1.6.4

Rewrite using the commutative property of multiplication.

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

Step 4.1.6.5

Cancel the common factor of $1 - 2 x$ .

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Step 4.1.6.5.1

Cancel the common factor.

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

Step 4.1.6.5.2

Divide $(B) (1 + x)$ by $1$ .

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

Step 4.1.6.6

Apply the distributive property.

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

Step 4.1.6.7

Multiply $B$ by $1$ .

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

Step 4.1.7

Simplify the expression.

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Step 4.1.7.1

Move $A$ .

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

Step 4.1.7.2

Reorder $B$ and $x$ .

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

Step 4.1.7.3

Move $B$ .

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

Step 4.1.7.4

Move $A$ .

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

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.

$0 = - 2 A + B$

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.

$1 = A + B$

Step 4.2.3

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

$0 = - 2 A + B$

$1 = A + B$

$0 = - 2 A + B$

$1 = A + B$

Step 4.3

Solve the system of equations.

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Step 4.3.1

Solve for $B$ in $0 = - 2 A + B$ .

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Step 4.3.1.1

Rewrite the equation as $- 2 A + B = 0$ .

$- 2 A + B = 0$

$1 = A + B$

Step 4.3.1.2

Add $2 A$ to both sides of the equation.

$B = 2 A$

$1 = A + B$

$B = 2 A$

$1 = A + B$

Step 4.3.2

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

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Step 4.3.2.1

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

$1 = A + 2 A$

$B = 2 A$

Step 4.3.2.2

Simplify $1 = A + 2 A$ .

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Step 4.3.2.2.1

Simplify the left side.

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Step 4.3.2.2.1.1

Remove parentheses.

$1 = A + 2 A$

$B = 2 A$

$1 = A + 2 A$

$B = 2 A$

Step 4.3.2.2.2

Simplify the right side.

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Step 4.3.2.2.2.1

Add $A$ and $2 A$ .

$1 = 3 A$

$B = 2 A$

$1 = 3 A$

$B = 2 A$

$1 = 3 A$

$B = 2 A$

$1 = 3 A$

$B = 2 A$

Step 4.3.3

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

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Step 4.3.3.1

Rewrite the equation as $3 A = 1$ .

$3 A = 1$

$B = 2 A$

Step 4.3.3.2

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

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Step 4.3.3.2.1

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

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

$B = 2 A$

Step 4.3.3.2.2

Simplify the left side.

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Step 4.3.3.2.2.1

Cancel the common factor of $3$ .

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Step 4.3.3.2.2.1.1

Cancel the common factor.

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

$B = 2 A$

Step 4.3.3.2.2.1.2

Divide $A$ by $1$ .

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

$B = 2 A$

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

$B = 2 A$

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

$B = 2 A$

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

$B = 2 A$

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

$B = 2 A$

Step 4.3.4

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

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Step 4.3.4.1

Replace all occurrences of $A$ in $B = 2 A$ with $\frac{1}{3}$ .

$B = 2 (\frac{1}{3})$

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

Step 4.3.4.2

Simplify the right side.

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Step 4.3.4.2.1

Combine $2$ and $\frac{1}{3}$ .

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

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

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

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

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

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

Step 4.3.5

List all of the solutions.

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

Step 4.4

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

$\frac{\frac{1}{3}}{1 + x} + \frac{\frac{2}{3}}{1 - 2 x}$

Step 4.5

Simplify.

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Step 4.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{3} \cdot \frac{1}{1 + x} + \frac{\frac{2}{3}}{1 - 2 x}$

Step 4.5.2

Multiply $\frac{1}{3}$ by $\frac{1}{1 + x}$ .

$\frac{1}{3 (1 + x)} + \frac{\frac{2}{3}}{1 - 2 x}$

Step 4.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{3 (1 + x)} + \frac{2}{3} \cdot \frac{1}{1 - 2 x}$

Step 4.5.4

Multiply $\frac{2}{3}$ by $\frac{1}{1 - 2 x}$ .

$\int \frac{1}{3 (1 + x)} + \frac{2}{3 (1 - 2 x)} d x$

Step 5

Split the single integral into multiple integrals.

$\int \frac{1}{3 (1 + x)} d x + \int \frac{2}{3 (1 - 2 x)} d x$

Step 6

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

$\frac{1}{3} \int \frac{1}{1 + x} d x + \int \frac{2}{3 (1 - 2 x)} d x$

Step 7

Let $u_{1} = 1 + x$ . Then $d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 7.1

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

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Step 7.1.1

Differentiate $1 + x$ .

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

Step 7.1.2

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

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

Step 7.1.3

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

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

Step 7.1.4

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

$0 + 1$

Step 7.1.5

Add $0$ and $1$ .

$1$

Step 7.2

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

$\frac{1}{3} \int \frac{1}{u_{1}} d u_{1} + \int \frac{2}{3 (1 - 2 x)} d x$

Step 8

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

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

Step 9

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

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

Step 10

Let $u_{2} = 1 - 2 x$ . 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 10.1

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

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Step 10.1.1

Differentiate $1 - 2 x$ .

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

Step 10.1.2

Differentiate.

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Step 10.1.2.1

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

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

Step 10.1.2.2

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

$0 + \frac{d}{d x} [- 2 x]$

Step 10.1.3

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

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Step 10.1.3.1

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

$0 - 2 \frac{d}{d x} [x]$

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 $n = 1$ .

$0 - 2 \cdot 1$

Step 10.1.3.3

Multiply $- 2$ by $1$ .

$0 - 2$

Step 10.1.4

Subtract $2$ from $0$ .

$- 2$

Step 10.2

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

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

Step 11

Simplify.

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Step 11.1

Move the negative in front of the fraction.

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

Step 11.2

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

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

Step 11.3

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

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

Step 12

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

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

Step 13

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

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

Step 14

Simplify.

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Step 14.1

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

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

Step 14.2

Multiply $3$ by $2$ .

$\frac{1}{3} (\ln (| u_{1} |) + C) - \frac{2}{6} \int \frac{1}{u_{2}} d u_{2}$

Step 14.3

Cancel the common factor of $2$ and $6$ .

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Step 14.3.1

Factor $2$ out of $2$ .

$\frac{1}{3} (\ln (| u_{1} |) + C) - \frac{2 (1)}{6} \int \frac{1}{u_{2}} d u_{2}$

Step 14.3.2

Cancel the common factors.

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Step 14.3.2.1

Factor $2$ out of $6$ .

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

Step 14.3.2.2

Cancel the common factor.

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

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 $1 + x$ .

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

Step 17.2

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

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

Step 18

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

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