Calculus Examples

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

Step 1

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

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

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.

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

Step 4

Write the fraction using partial fraction decomposition.

Tap for more steps...

Step 4.1

Decompose the fraction and multiply through by the common denominator.

Tap for more steps...

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 $A$ .

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

Step 4.1.2

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

Step 4.1.3

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

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

Step 4.1.4

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

Tap for more steps...

Step 4.1.4.1

Cancel the common factor.

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

Step 4.1.4.2

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

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

Step 4.1.5

Simplify each term.

Tap for more steps...

Step 4.1.5.1

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

Tap for more steps...

Step 4.1.5.1.1

Cancel the common factor.

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

Step 4.1.5.1.2

Divide $A$ by $1$ .

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

Step 4.1.5.2

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

Tap for more steps...

Step 4.1.5.2.1

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

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

Step 4.1.5.2.2

Cancel the common factors.

Tap for more steps...

Step 4.1.5.2.2.1

Multiply by $1$ .

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

Step 4.1.5.2.2.2

Cancel the common factor.

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

Step 4.1.5.2.2.3

Rewrite the expression.

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

Step 4.1.5.2.2.4

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

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

Step 4.1.5.3

Apply the distributive property.

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

Step 4.1.5.4

Multiply $B$ by $1$ .

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

Step 4.1.6

Reorder $A$ and $B x$ .

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

Step 4.2

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

Tap for more steps...

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.

$2 = B$

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.

$2 = B$

$- 1 = A + B$

$2 = B$

$- 1 = A + B$

Step 4.3

Solve the system of equations.

Tap for more steps...

Step 4.3.1

Rewrite the equation as $B = 2$ .

$B = 2$

$- 1 = A + B$

Step 4.3.2

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

Tap for more steps...

Step 4.3.2.1

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

$- 1 = A + 2$

$B = 2$

Step 4.3.2.2

Simplify the left side.

Tap for more steps...

Step 4.3.2.2.1

Remove parentheses.

$- 1 = A + 2$

$B = 2$

$- 1 = A + 2$

$B = 2$

$- 1 = A + 2$

$B = 2$

Step 4.3.3

Solve for $A$ in $- 1 = A + 2$ .

Tap for more steps...

Step 4.3.3.1

Rewrite the equation as $A + 2 = - 1$ .

$A + 2 = - 1$

$B = 2$

Step 4.3.3.2

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

Tap for more steps...

Step 4.3.3.2.1

Subtract $2$ from both sides of the equation.

$A = - 1 - 2$

$B = 2$

Step 4.3.3.2.2

Subtract $2$ from $- 1$ .

$A = - 3$

$B = 2$

$A = - 3$

$B = 2$

$A = - 3$

$B = 2$

Step 4.3.4

Solve the system of equations.

$A = - 3 B = 2$

Step 4.3.5

List all of the solutions.

$A = - 3, B = 2$

Step 4.4

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

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

Step 4.5

Move the negative in front of the fraction.

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

Step 5

Split the single integral into multiple integrals.

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

Step 6

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

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

Step 7

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

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

Step 8

Multiply $3$ by $- 1$ .

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

Step 9

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

Tap for more steps...

Step 9.1

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

Tap for more steps...

Step 9.1.1

Differentiate $x + 1$ .

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

Step 9.1.2

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 9.1.3

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 9.1.4

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

$1 + 0$

Step 9.1.5

Add $1$ and $0$ .

$1$

Step 9.2

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

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

Step 10

Apply basic rules of exponents.

Tap for more steps...

Step 10.1

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

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

Step 10.2

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

Tap for more steps...

Step 10.2.1

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

$- 3 \int {u_{1}}^{2 \cdot - 1} d u_{1} + \int \frac{2}{x + 1} d x$

Step 10.2.2

Multiply $2$ by $- 1$ .

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

Step 11

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

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

Step 12

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

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

Step 13

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

Tap for more steps...

Step 13.1

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

Tap for more steps...

Step 13.1.1

Differentiate $x + 1$ .

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

Step 13.1.2

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 13.1.3

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 13.1.4

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

$1 + 0$

Step 13.1.5

Add $1$ and $0$ .

$1$

Step 13.2

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

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

Step 14

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

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

Step 15

Simplify.

Tap for more steps...

Step 15.1

Simplify.

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

Step 15.2

Simplify.

Tap for more steps...

Step 15.2.1

Multiply $- 1$ by $- 3$ .

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

Step 15.2.2

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

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

Step 16

Substitute back in for each integration substitution variable.

Tap for more steps...

Step 16.1

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

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

Step 16.2

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

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

Step 17

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

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