Calculus Examples

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

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

Factor $x$ out of $x - x^{2}$ .

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Step 1.1.1.1

Raise $x$ to the power of $1$ .

$\frac{1}{x^{1} - x^{2}}$

Step 1.1.1.2

Factor $x$ out of $x^{1}$ .

$\frac{1}{x \cdot 1 - x^{2}}$

Step 1.1.1.3

Factor $x$ out of $- x^{2}$ .

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

Step 1.1.1.4

Factor $x$ out of $x \cdot 1 + x (- x)$ .

$\frac{1}{x \cdot (1 - x)}$

Step 1.1.1.5

Multiply $x$ by $1 - x$ .

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

Step 1.1.2

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 $B$ .

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

Step 1.1.3

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

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

Step 1.1.4

Cancel the common factor of $x$ .

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Step 1.1.4.1

Cancel the common factor.

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

Step 1.1.4.2

Rewrite the expression.

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

Step 1.1.5

Cancel the common factor of $1 - x$ .

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Step 1.1.5.1

Cancel the common factor.

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

Step 1.1.5.2

Rewrite the expression.

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

Step 1.1.6

Simplify each term.

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Step 1.1.6.1

Cancel the common factor of $x$ .

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Step 1.1.6.1.1

Cancel the common factor.

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

Step 1.1.6.1.2

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

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

Step 1.1.6.2

Apply the distributive property.

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

Step 1.1.6.3

Multiply $A$ by $1$ .

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

Step 1.1.6.4

Rewrite using the commutative property of multiplication.

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

Step 1.1.6.5

Cancel the common factor of $1 - x$ .

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Step 1.1.6.5.1

Cancel the common factor.

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

Step 1.1.6.5.2

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

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

$1 = A - A x + B x$

Step 1.1.7

Simplify the expression.

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Step 1.1.7.1

Move $A$ .

$1 = A - x A + B x$

Step 1.1.7.2

Reorder $B$ and $x$ .

$1 = A - x A + B x$

Step 1.1.7.3

Move $A$ .

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

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.

$0 = - 1 A + B$

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.

$1 = A$

Step 1.2.3

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

$0 = - 1 A + B$

$1 = A$

$0 = - 1 A + B$

$1 = A$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Rewrite the equation as $A = 1$ .

$A = 1$

$0 = - 1 A + B$

Step 1.3.2

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

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Step 1.3.2.1

Replace all occurrences of $A$ in $0 = - 1 A + B$ with $1$ .

$0 = - 1 \cdot 1 + B$

$A = 1$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Multiply $- 1$ by $1$ .

$0 = - 1 + B$

$A = 1$

$0 = - 1 + B$

$A = 1$

$0 = - 1 + B$

$A = 1$

Step 1.3.3

Solve for $B$ in $0 = - 1 + B$ .

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Step 1.3.3.1

Rewrite the equation as $- 1 + B = 0$ .

$- 1 + B = 0$

$A = 1$

Step 1.3.3.2

Add $1$ to both sides of the equation.

$B = 1$

$A = 1$

$B = 1$

$A = 1$

Step 1.3.4

Solve the system of equations.

$B = 1 A = 1$

Step 1.3.5

List all of the solutions.

$B = 1, A = 1$

Step 1.4

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

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

Step 1.5

Remove the zero from the expression.

$\int \frac{1}{x} + \frac{1}{1 - x} d x$

Step 2

Split the single integral into multiple integrals.

$\int \frac{1}{x} d x + \int \frac{1}{1 - x} d x$

Step 3

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

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

Step 4

Let $u = 1 - x$ . Then $d u = - d x$ , so $- d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.1

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

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Step 4.1.1

Rewrite.

$\frac{- 1}{1}$

Step 4.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 4.2

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

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

Step 5

Move the negative in front of the fraction.

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

Step 6

Since $- 1$ is constant with respect to $u$ , move $- 1$ out of the integral.

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

Step 7

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

$\ln (| x |) + C - (\ln (| u |) + C)$

Step 8

Simplify.

$\ln (| x |) - \ln (| u |) + C$

Step 9

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{| x |}{| u |}) + C$

Step 10

Replace all occurrences of $u$ with $1 - x$ .

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