Calculus Examples

$\frac{d y}{d x} = \frac{1 - 2 x}{4 x - x^{2}}$

Step 1

Rewrite the equation.

$d y = \frac{1 - 2 x}{4 x - x^{2}} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int \frac{1 - 2 x}{4 x - x^{2}} d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int \frac{1 - 2 x}{4 x - x^{2}} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Write the fraction using partial fraction decomposition.

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Step 2.3.1.1

Decompose the fraction and multiply through by the common denominator.

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Step 2.3.1.1.1

Factor $x$ out of $4 x - x^{2}$ .

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Step 2.3.1.1.1.1

Factor $x$ out of $4 x$ .

$\frac{1 - 2 x}{x \cdot 4 - x^{2}}$

Step 2.3.1.1.1.2

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

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

Step 2.3.1.1.1.3

Factor $x$ out of $x \cdot 4 + x (- x)$ .

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

Step 2.3.1.1.1.4

Multiply $x$ by $4 - x$ .

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

Step 2.3.1.1.2

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 $B$ .

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

Step 2.3.1.1.3

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

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

Step 2.3.1.1.4

Cancel the common factor of $x$ .

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Step 2.3.1.1.4.1

Cancel the common factor.

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

Step 2.3.1.1.4.2

Rewrite the expression.

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

Step 2.3.1.1.5

Cancel the common factor of $4 - x$ .

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Step 2.3.1.1.5.1

Cancel the common factor.

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

Step 2.3.1.1.5.2

Divide $1 - 2 x$ by $1$ .

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

Step 2.3.1.1.6

Reorder $1$ and $- 2 x$ .

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

Step 2.3.1.1.7

Simplify each term.

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Step 2.3.1.1.7.1

Cancel the common factor of $x$ .

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Step 2.3.1.1.7.1.1

Cancel the common factor.

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

Step 2.3.1.1.7.1.2

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

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

Step 2.3.1.1.7.2

Apply the distributive property.

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

Step 2.3.1.1.7.3

Move $4$ to the left of $A$ .

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

Step 2.3.1.1.7.4

Rewrite using the commutative property of multiplication.

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

Step 2.3.1.1.7.5

Cancel the common factor of $4 - x$ .

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Step 2.3.1.1.7.5.1

Cancel the common factor.

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

Step 2.3.1.1.7.5.2

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

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

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

Step 2.3.1.1.8

Simplify the expression.

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Step 2.3.1.1.8.1

Move $A$ .

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

Step 2.3.1.1.8.2

Reorder $B$ and $x$ .

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

Step 2.3.1.1.8.3

Move $4 A$ .

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

Step 2.3.1.2

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

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Step 2.3.1.2.1

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.

$- 2 = - 1 A + B$

Step 2.3.1.2.2

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.

$1 = 4 A$

Step 2.3.1.2.3

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

$- 2 = - 1 A + B$

$1 = 4 A$

$- 2 = - 1 A + B$

$1 = 4 A$

Step 2.3.1.3

Solve the system of equations.

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Step 2.3.1.3.1

Solve for $A$ in $1 = 4 A$ .

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Step 2.3.1.3.1.1

Rewrite the equation as $4 A = 1$ .

$4 A = 1$

$- 2 = - 1 A + B$

Step 2.3.1.3.1.2

Divide each term in $4 A = 1$ by $4$ and simplify.

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Step 2.3.1.3.1.2.1

Divide each term in $4 A = 1$ by $4$ .

$\frac{4 A}{4} = \frac{1}{4}$

$- 2 = - 1 A + B$

Step 2.3.1.3.1.2.2

Simplify the left side.

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Step 2.3.1.3.1.2.2.1

Cancel the common factor of $4$ .

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Step 2.3.1.3.1.2.2.1.1

Cancel the common factor.

$\frac{4 A}{4} = \frac{1}{4}$

$- 2 = - 1 A + B$

Step 2.3.1.3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{1}{4}$

$- 2 = - 1 A + B$

$A = \frac{1}{4}$

$- 2 = - 1 A + B$

$A = \frac{1}{4}$

$- 2 = - 1 A + B$

$A = \frac{1}{4}$

$- 2 = - 1 A + B$

$A = \frac{1}{4}$

$- 2 = - 1 A + B$

Step 2.3.1.3.2

Replace all occurrences of $A$ with $\frac{1}{4}$ in each equation.

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Step 2.3.1.3.2.1

Replace all occurrences of $A$ in $- 2 = - 1 A + B$ with $\frac{1}{4}$ .

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

$A = \frac{1}{4}$

Step 2.3.1.3.2.2

Simplify the right side.

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Step 2.3.1.3.2.2.1

Rewrite $- 1 (\frac{1}{4})$ as $- (\frac{1}{4})$ .

$- 2 = - \frac{1}{4} + B$

$A = \frac{1}{4}$

$- 2 = - \frac{1}{4} + B$

$A = \frac{1}{4}$

$- 2 = - \frac{1}{4} + B$

$A = \frac{1}{4}$

Step 2.3.1.3.3

Solve for $B$ in $- 2 = - \frac{1}{4} + B$ .

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Step 2.3.1.3.3.1

Rewrite the equation as $- \frac{1}{4} + B = - 2$ .

$- \frac{1}{4} + B = - 2$

$A = \frac{1}{4}$

Step 2.3.1.3.3.2

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

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Step 2.3.1.3.3.2.1

Add $\frac{1}{4}$ to both sides of the equation.

$B = - 2 + \frac{1}{4}$

$A = \frac{1}{4}$

Step 2.3.1.3.3.2.2

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$B = - 2 \cdot \frac{4}{4} + \frac{1}{4}$

$A = \frac{1}{4}$

Step 2.3.1.3.3.2.3

Combine $- 2$ and $\frac{4}{4}$ .

$B = \frac{- 2 \cdot 4}{4} + \frac{1}{4}$

$A = \frac{1}{4}$

Step 2.3.1.3.3.2.4

Combine the numerators over the common denominator.

$B = \frac{- 2 \cdot 4 + 1}{4}$

$A = \frac{1}{4}$

Step 2.3.1.3.3.2.5

Simplify the numerator.

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Step 2.3.1.3.3.2.5.1

Multiply $- 2$ by $4$ .

$B = \frac{- 8 + 1}{4}$

$A = \frac{1}{4}$

Step 2.3.1.3.3.2.5.2

Add $- 8$ and $1$ .

$B = \frac{- 7}{4}$

$A = \frac{1}{4}$

$B = \frac{- 7}{4}$

$A = \frac{1}{4}$

Step 2.3.1.3.3.2.6

Move the negative in front of the fraction.

$B = - \frac{7}{4}$

$A = \frac{1}{4}$

$B = - \frac{7}{4}$

$A = \frac{1}{4}$

$B = - \frac{7}{4}$

$A = \frac{1}{4}$

Step 2.3.1.3.4

Solve the system of equations.

$B = - \frac{7}{4} A = \frac{1}{4}$

Step 2.3.1.3.5

List all of the solutions.

$B = - \frac{7}{4}, A = \frac{1}{4}$

Step 2.3.1.4

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

$\frac{\frac{1}{4}}{x} + \frac{- \frac{7}{4}}{4 - x}$

Step 2.3.1.5

Simplify.

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Step 2.3.1.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{4} \cdot \frac{1}{x} + \frac{- \frac{7}{4}}{4 - x}$

Step 2.3.1.5.2

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

$\frac{1}{4 x} + \frac{- \frac{7}{4}}{4 - x}$

Step 2.3.1.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{4 x} - \frac{7}{4} \cdot \frac{1}{4 - x}$

Step 2.3.1.5.4

Multiply $\frac{1}{4 - x}$ by $\frac{7}{4}$ .

$\frac{1}{4 x} - \frac{7}{(4 - x) \cdot 4}$

Step 2.3.1.5.5

Move $4$ to the left of $4 - x$ .

$y + C_{1} = \int \frac{1}{4 x} - \frac{7}{4 (4 - x)} d x$

Step 2.3.2

Split the single integral into multiple integrals.

$y + C_{1} = \int \frac{1}{4 x} d x + \int - \frac{7}{4 (4 - x)} d x$

Step 2.3.3

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

$y + C_{1} = \frac{1}{4} \int \frac{1}{x} d x + \int - \frac{7}{4 (4 - x)} d x$

Step 2.3.4

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

$y + C_{1} = \frac{1}{4} (\ln (| x |) + C_{2}) + \int - \frac{7}{4 (4 - x)} d x$

Step 2.3.5

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

$y + C_{1} = \frac{1}{4} (\ln (| x |) + C_{2}) - \int \frac{7}{4 (4 - x)} d x$

Step 2.3.6

Since $\frac{7}{4}$ is constant with respect to $x$ , move $\frac{7}{4}$ out of the integral.

$y + C_{1} = \frac{1}{4} (\ln (| x |) + C_{2}) - (\frac{7}{4} \int \frac{1}{4 - x} d x)$

Step 2.3.7

Let $u = 4 - x$ . Then $d u = - d x$ , so $- d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.7.1

Let $u = 4 - x$ . Find $\frac{d u}{d x}$ .

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Step 2.3.7.1.1

Rewrite.

$\frac{- 1}{1}$

Step 2.3.7.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 2.3.7.2

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

$y + C_{1} = \frac{1}{4} (\ln (| x |) + C_{2}) - \frac{7}{4} \int \frac{- 1}{u} d u$

Step 2.3.8

Move the negative in front of the fraction.

$y + C_{1} = \frac{1}{4} (\ln (| x |) + C_{2}) - \frac{7}{4} \int - \frac{1}{u} d u$

Step 2.3.9

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

$y + C_{1} = \frac{1}{4} (\ln (| x |) + C_{2}) - \frac{7}{4} (- \int \frac{1}{u} d u)$

Step 2.3.10

Simplify.

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Step 2.3.10.1

Multiply $- 1$ by $- 1$ .

$y + C_{1} = \frac{1}{4} (\ln (| x |) + C_{2}) + 1 (\frac{7}{4}) \int \frac{1}{u} d u$

Step 2.3.10.2

Multiply $\frac{7}{4}$ by $1$ .

$y + C_{1} = \frac{1}{4} (\ln (| x |) + C_{2}) + \frac{7}{4} \int \frac{1}{u} d u$

Step 2.3.11

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

$y + C_{1} = \frac{1}{4} (\ln (| x |) + C_{2}) + \frac{7}{4} (\ln (| u |) + C_{3})$

Step 2.3.12

Simplify.

$y + C_{1} = \frac{1}{4} \ln (| x |) + \frac{7}{4} \ln (| u |) + C_{4}$

Step 2.3.13

Replace all occurrences of $u$ with $4 - x$ .

$y + C_{1} = \frac{1}{4} \ln (| x |) + \frac{7}{4} \ln (| 4 - x |) + C_{4}$

Step 2.4

Group the constant of integration on the right side as $C$ .

$y = \frac{1}{4} \ln (| x |) + \frac{7}{4} \ln (| 4 - x |) + C$