Calculus Examples

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

Step 1

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

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

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.

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

Step 4

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

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

Step 5

Write the fraction using partial fraction decomposition.

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Step 5.1

Decompose the fraction and multiply through by the common denominator.

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Step 5.1.1

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 $A$ .

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

Step 5.1.2

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

Step 5.1.3

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

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

Step 5.1.4

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

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Step 5.1.4.1

Cancel the common factor.

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

Step 5.1.4.2

Divide $x$ by $1$ .

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

Step 5.1.5

Simplify each term.

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Step 5.1.5.1

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

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Step 5.1.5.1.1

Cancel the common factor.

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

Step 5.1.5.1.2

Divide $A$ by $1$ .

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

Step 5.1.5.2

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

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Step 5.1.5.2.1

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

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

Step 5.1.5.2.2

Cancel the common factors.

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Step 5.1.5.2.2.1

Multiply by $1$ .

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

Step 5.1.5.2.2.2

Cancel the common factor.

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

Step 5.1.5.2.2.3

Rewrite the expression.

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

Step 5.1.5.2.2.4

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

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

Step 5.1.5.3

Apply the distributive property.

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

Step 5.1.5.4

Multiply $B$ by $1$ .

$x = A + B x + B$

Step 5.1.6

Reorder $A$ and $B x$ .

$x = B x + A + B$

Step 5.2

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

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Step 5.2.1

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.

$1 = B$

Step 5.2.2

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 5.2.3

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

$1 = B$

$0 = A + B$

$1 = B$

$0 = A + B$

Step 5.3

Solve the system of equations.

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Step 5.3.1

Rewrite the equation as $B = 1$ .

$B = 1$

$0 = A + B$

Step 5.3.2

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

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Step 5.3.2.1

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

$0 = A + 1$

$B = 1$

Step 5.3.2.2

Simplify the left side.

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Step 5.3.2.2.1

Remove parentheses.

$0 = A + 1$

$B = 1$

$0 = A + 1$

$B = 1$

$0 = A + 1$

$B = 1$

Step 5.3.3

Solve for $A$ in $0 = A + 1$ .

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Step 5.3.3.1

Rewrite the equation as $A + 1 = 0$ .

$A + 1 = 0$

$B = 1$

Step 5.3.3.2

Subtract $1$ from both sides of the equation.

$A = - 1$

$B = 1$

$A = - 1$

$B = 1$

Step 5.3.4

Solve the system of equations.

$A = - 1 B = 1$

Step 5.3.5

List all of the solutions.

$A = - 1, B = 1$

Step 5.4

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

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

Step 5.5

Move the negative in front of the fraction.

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

Step 6

Split the single integral into multiple integrals.

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

Step 7

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

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

Step 8

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

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Step 8.1

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

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Step 8.1.1

Differentiate $x + 1$ .

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

Step 8.1.2

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

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

Step 8.1.3

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

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

Step 8.1.4

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

$1 + 0$

Step 8.1.5

Add $1$ and $0$ .

$1$

Step 8.2

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

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

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.

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

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, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 9.2.2

Multiply $2$ by $- 1$ .

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

Step 10

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

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

Step 11

Let $u_{2} = x + 1$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 11.1

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

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Step 11.1.1

Differentiate $x + 1$ .

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

Step 11.1.2

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

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

Step 11.1.3

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

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

Step 11.1.4

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

$1 + 0$

Step 11.1.5

Add $1$ and $0$ .

$1$

Step 11.2

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

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

Step 12

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

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

Step 13

Simplify.

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

Step 14

Substitute back in for each integration substitution variable.

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Step 14.1

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

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

Step 14.2

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

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

Step 15

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

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