Calculus Examples

$\int \frac{x}{{(1 + 4 x)}^{2}} d x$

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

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 $A$ .

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

Step 1.1.2

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

Step 1.1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is ${(1 + 4 x)}^{2}$ .

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

Step 1.1.4

Cancel the common factor of ${(1 + 4 x)}^{2}$ .

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Step 1.1.4.1

Cancel the common factor.

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

Step 1.1.4.2

Divide $x$ by $1$ .

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

Step 1.1.5

Simplify each term.

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Step 1.1.5.1

Cancel the common factor of ${(1 + 4 x)}^{2}$ .

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Step 1.1.5.1.1

Cancel the common factor.

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

Step 1.1.5.1.2

Divide $A$ by $1$ .

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

Step 1.1.5.2

Cancel the common factor of ${(1 + 4 x)}^{2}$ and $1 + 4 x$ .

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Step 1.1.5.2.1

Factor $1 + 4 x$ out of $(B) {(1 + 4 x)}^{2}$ .

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

Step 1.1.5.2.2

Cancel the common factors.

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Step 1.1.5.2.2.1

Multiply by $1$ .

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

Step 1.1.5.2.2.2

Cancel the common factor.

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

Step 1.1.5.2.2.3

Rewrite the expression.

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

Step 1.1.5.2.2.4

Divide $B (1 + 4 x)$ by $1$ .

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

Step 1.1.5.3

Apply the distributive property.

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

Step 1.1.5.4

Multiply $B$ by $1$ .

$x = A + B + B (4 x)$

Step 1.1.5.5

Rewrite using the commutative property of multiplication.

$x = A + B + 4 B x$

Step 1.1.6

Simplify the expression.

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Step 1.1.6.1

Move $B$ .

$x = A + B + 4 x B$

Step 1.1.6.2

Move $B$ .

$x = A + 4 x B + B$

Step 1.1.6.3

Reorder $A$ and $4 x B$ .

$x = 4 x B + A + B$

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.

$1 = 4 B$

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.

$0 = A + B$

Step 1.2.3

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

$1 = 4 B$

$0 = A + B$

$1 = 4 B$

$0 = A + B$

Step 1.3

Solve the system of equations.

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Step 1.3.1

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

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Step 1.3.1.1

Rewrite the equation as $4 B = 1$ .

$4 B = 1$

$0 = A + B$

Step 1.3.1.2

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

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Step 1.3.1.2.1

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

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

$0 = A + B$

Step 1.3.1.2.2

Simplify the left side.

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Step 1.3.1.2.2.1

Cancel the common factor of $4$ .

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Step 1.3.1.2.2.1.1

Cancel the common factor.

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

$0 = A + B$

Step 1.3.1.2.2.1.2

Divide $B$ by $1$ .

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

$0 = A + B$

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

$0 = A + B$

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

$0 = A + B$

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

$0 = A + B$

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

$0 = A + B$

Step 1.3.2

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

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Step 1.3.2.1

Replace all occurrences of $B$ in $0 = A + B$ with $\frac{1}{4}$ .

$0 = A + \frac{1}{4}$

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

Step 1.3.2.2

Simplify the left side.

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Step 1.3.2.2.1

Remove parentheses.

$0 = A + \frac{1}{4}$

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

$0 = A + \frac{1}{4}$

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

$0 = A + \frac{1}{4}$

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

Step 1.3.3

Solve for $A$ in $0 = A + \frac{1}{4}$ .

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Step 1.3.3.1

Rewrite the equation as $A + \frac{1}{4} = 0$ .

$A + \frac{1}{4} = 0$

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

Step 1.3.3.2

Subtract $\frac{1}{4}$ from both sides of the equation.

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

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

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

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

Step 1.3.4

Solve the system of equations.

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

Step 1.3.5

List all of the solutions.

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

Step 1.4

Replace each of the partial fraction coefficients in $\frac{A}{{(1 + 4 x)}^{2}} + \frac{B}{1 + 4 x}$ with the values found for $A$ and $B$ .

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

Step 1.5

Simplify.

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Step 1.5.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.5.2

Multiply $\frac{1}{{(1 + 4 x)}^{2}}$ by $\frac{1}{4}$ .

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

Step 1.5.3

Move $4$ to the left of ${(1 + 4 x)}^{2}$ .

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

Step 1.5.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.5.5

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

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

Step 2

Split the single integral into multiple integrals.

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

Step 3

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

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

Step 4

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

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

Step 5

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

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Step 5.1

Let $u_{1} = 1 + 4 x$ . Find $\frac{d u_{1}}{d x}$ .

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Step 5.1.1

Differentiate $1 + 4 x$ .

$\frac{d}{d x} [1 + 4 x]$

Step 5.1.2

Differentiate.

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Step 5.1.2.1

By the Sum Rule, the derivative of $1 + 4 x$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [4 x]$ .

$\frac{d}{d x} [1] + \frac{d}{d x} [4 x]$

Step 5.1.2.2

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

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

Step 5.1.3

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

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Step 5.1.3.1

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 5.1.3.2

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 5.1.3.3

Multiply $4$ by $1$ .

$0 + 4$

Step 5.1.4

Add $0$ and $4$ .

$4$

Step 5.2

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

$- \frac{1}{4} \int \frac{1}{{u_{1}}^{2}} \cdot \frac{1}{4} d u_{1} + \int \frac{1}{4 (1 + 4 x)} d x$

Step 6

Simplify.

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Step 6.1

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

$- \frac{1}{4} \int \frac{1}{{u_{1}}^{2} \cdot 4} d u_{1} + \int \frac{1}{4 (1 + 4 x)} d x$

Step 6.2

Move $4$ to the left of ${u_{1}}^{2}$ .

$- \frac{1}{4} \int \frac{1}{4 {u_{1}}^{2}} d u_{1} + \int \frac{1}{4 (1 + 4 x)} d x$

Step 7

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

$- \frac{1}{4} (\frac{1}{4} \int \frac{1}{{u_{1}}^{2}} d u_{1}) + \int \frac{1}{4 (1 + 4 x)} d x$

Step 8

Simplify the expression.

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Step 8.1

Simplify.

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Step 8.1.1

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

$- \frac{1}{4 \cdot 4} \int \frac{1}{{u_{1}}^{2}} d u_{1} + \int \frac{1}{4 (1 + 4 x)} d x$

Step 8.1.2

Multiply $4$ by $4$ .

$- \frac{1}{16} \int \frac{1}{{u_{1}}^{2}} d u_{1} + \int \frac{1}{4 (1 + 4 x)} d x$

Step 8.2

Apply basic rules of exponents.

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Step 8.2.1

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

$- \frac{1}{16} \int {({u_{1}}^{2})}^{- 1} d u_{1} + \int \frac{1}{4 (1 + 4 x)} d x$

Step 8.2.2

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

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Step 8.2.2.1

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

$- \frac{1}{16} \int {u_{1}}^{2 \cdot - 1} d u_{1} + \int \frac{1}{4 (1 + 4 x)} d x$

Step 8.2.2.2

Multiply $2$ by $- 1$ .

$- \frac{1}{16} \int {u_{1}}^{- 2} d u_{1} + \int \frac{1}{4 (1 + 4 x)} d x$

Step 9

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

$- \frac{1}{16} (- {u_{1}}^{- 1} + C) + \int \frac{1}{4 (1 + 4 x)} d x$

Step 10

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

$- \frac{1}{16} (- {u_{1}}^{- 1} + C) + \frac{1}{4} \int \frac{1}{1 + 4 x} d x$

Step 11

Let $u_{2} = 1 + 4 x$ . 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 11.1

Let $u_{2} = 1 + 4 x$ . Find $\frac{d u_{2}}{d x}$ .

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Step 11.1.1

Differentiate $1 + 4 x$ .

$\frac{d}{d x} [1 + 4 x]$

Step 11.1.2

Differentiate.

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Step 11.1.2.1

By the Sum Rule, the derivative of $1 + 4 x$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [4 x]$ .

$\frac{d}{d x} [1] + \frac{d}{d x} [4 x]$

Step 11.1.2.2

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

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

Step 11.1.3

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

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Step 11.1.3.1

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 11.1.3.2

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 11.1.3.3

Multiply $4$ by $1$ .

$0 + 4$

Step 11.1.4

Add $0$ and $4$ .

$4$

Step 11.2

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

$- \frac{1}{16} (- {u_{1}}^{- 1} + C) + \frac{1}{4} \int \frac{1}{u_{2}} \cdot \frac{1}{4} d u_{2}$

Step 12

Simplify.

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Step 12.1

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

$- \frac{1}{16} (- {u_{1}}^{- 1} + C) + \frac{1}{4} \int \frac{1}{u_{2} \cdot 4} d u_{2}$

Step 12.2

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

$- \frac{1}{16} (- {u_{1}}^{- 1} + C) + \frac{1}{4} \int \frac{1}{4 u_{2}} d u_{2}$

Step 13

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

$- \frac{1}{16} (- {u_{1}}^{- 1} + C) + \frac{1}{4} (\frac{1}{4} \int \frac{1}{u_{2}} d u_{2})$

Step 14

Simplify.

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Step 14.1

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

$- \frac{1}{16} (- {u_{1}}^{- 1} + C) + \frac{1}{4 \cdot 4} \int \frac{1}{u_{2}} d u_{2}$

Step 14.2

Multiply $4$ by $4$ .

$- \frac{1}{16} (- {u_{1}}^{- 1} + C) + \frac{1}{16} \int \frac{1}{u_{2}} d u_{2}$

Step 15

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

$- \frac{1}{16} (- {u_{1}}^{- 1} + C) + \frac{1}{16} (\ln (| u_{2} |) + C)$

Step 16

Simplify.

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Step 16.1

Simplify.

$- \frac{1}{16} (- \frac{1}{u_{1}}) + \frac{1}{16} \ln (| u_{2} |) + C$

Step 16.2

Simplify.

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Step 16.2.1

Multiply $- 1$ by $- 1$ .

$1 (\frac{1}{16}) \frac{1}{u_{1}} + \frac{1}{16} \ln (| u_{2} |) + C$

Step 16.2.2

Multiply $\frac{1}{16}$ by $1$ .

$\frac{1}{16} \cdot \frac{1}{u_{1}} + \frac{1}{16} \ln (| u_{2} |) + C$

Step 16.2.3

Multiply $\frac{1}{16}$ by $\frac{1}{u_{1}}$ .

$\frac{1}{16 u_{1}} + \frac{1}{16} \ln (| u_{2} |) + C$

Step 17

Substitute back in for each integration substitution variable.

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Step 17.1

Replace all occurrences of $u_{1}$ with $1 + 4 x$ .

$\frac{1}{16 (1 + 4 x)} + \frac{1}{16} \ln (| u_{2} |) + C$

Step 17.2

Replace all occurrences of $u_{2}$ with $1 + 4 x$ .

$\frac{1}{16 (1 + 4 x)} + \frac{1}{16} \ln (| 1 + 4 x |) + C$