Calculus Examples

$(2 x + 3) d x + (x^{2} - 1) d y = 0$

Step 1

Subtract $(2 x + 3) d x$ from both sides of the equation.

$(x^{2} - 1) d y = - (2 x + 3) d x$

Step 2

Multiply both sides by $\frac{1}{x^{2} - 1}$ .

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

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $x^{2} - 1$ .

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Step 3.1.1

Cancel the common factor.

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

Step 3.1.2

Rewrite the expression.

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

Step 3.2

Rewrite using the commutative property of multiplication.

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

Step 3.3

Simplify the denominator.

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Step 3.3.1

Rewrite $1$ as $1^{2}$ .

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

Step 3.3.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$1 d y = - \frac{1}{(x + 1) (x - 1)} (2 x + 3) d x$

Step 3.4

Apply the distributive property.

$1 d y = (- \frac{1}{(x + 1) (x - 1)} (2 x) - \frac{1}{(x + 1) (x - 1)} \cdot 3) d x$

Step 3.5

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

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Step 3.5.1

Multiply $2$ by $- 1$ .

$1 d y = (- 2 \frac{1}{(x + 1) (x - 1)} x - \frac{1}{(x + 1) (x - 1)} \cdot 3) d x$

Step 3.5.2

Combine $- 2$ and $\frac{1}{(x + 1) (x - 1)}$ .

$1 d y = (\frac{- 2}{(x + 1) (x - 1)} x - \frac{1}{(x + 1) (x - 1)} \cdot 3) d x$

Step 3.5.3

Combine $\frac{- 2}{(x + 1) (x - 1)}$ and $x$ .

$1 d y = (\frac{- 2 x}{(x + 1) (x - 1)} - \frac{1}{(x + 1) (x - 1)} \cdot 3) d x$

Step 3.6

Multiply $- \frac{1}{(x + 1) (x - 1)} \cdot 3$ .

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Step 3.6.1

Multiply $3$ by $- 1$ .

$1 d y = (\frac{- 2 x}{(x + 1) (x - 1)} - 3 \frac{1}{(x + 1) (x - 1)}) d x$

Step 3.6.2

Combine $- 3$ and $\frac{1}{(x + 1) (x - 1)}$ .

$1 d y = (\frac{- 2 x}{(x + 1) (x - 1)} + \frac{- 3}{(x + 1) (x - 1)}) d x$

Step 3.7

Simplify each term.

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Step 3.7.1

Move the negative in front of the fraction.

$1 d y = (- \frac{2 x}{(x + 1) (x - 1)} + \frac{- 3}{(x + 1) (x - 1)}) d x$

Step 3.7.2

Move the negative in front of the fraction.

$1 d y = (- \frac{2 x}{(x + 1) (x - 1)} - \frac{3}{(x + 1) (x - 1)}) d x$

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Apply the constant rule.

$y + C_{1} = \int - \frac{2 x}{(x + 1) (x - 1)} - \frac{3}{(x + 1) (x - 1)} d x$

Step 4.3

Integrate the right side.

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Step 4.3.1

Split the single integral into multiple integrals.

$y + C_{1} = \int - \frac{2 x}{(x + 1) (x - 1)} d x + \int - \frac{3}{(x + 1) (x - 1)} d x$

Step 4.3.2

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

$y + C_{1} = - \int \frac{2 x}{(x + 1) (x - 1)} d x + \int - \frac{3}{(x + 1) (x - 1)} d x$

Step 4.3.3

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

$y + C_{1} = - (2 \int \frac{x}{(x + 1) (x - 1)} d x) + \int - \frac{3}{(x + 1) (x - 1)} d x$

Step 4.3.4

Multiply $2$ by $- 1$ .

$y + C_{1} = - 2 \int \frac{x}{(x + 1) (x - 1)} d x + \int - \frac{3}{(x + 1) (x - 1)} d x$

Step 4.3.5

Let $u_{1} = (x + 1) (x - 1)$ . Then $d u_{1} = 2 x d x$ , so $\frac{1}{2} d u_{1} = x d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 4.3.5.1

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

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Step 4.3.5.1.1

Differentiate $(x + 1) (x - 1)$ .

$\frac{d}{d x} [(x + 1) (x - 1)]$

Step 4.3.5.1.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x + 1$ and $g (x) = x - 1$ .

$(x + 1) \frac{d}{d x} [x - 1] + (x - 1) \frac{d}{d x} [x + 1]$

Step 4.3.5.1.3

Differentiate.

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Step 4.3.5.1.3.1

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

$(x + 1) (\frac{d}{d x} [x] + \frac{d}{d x} [- 1]) + (x - 1) \frac{d}{d x} [x + 1]$

Step 4.3.5.1.3.2

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

$(x + 1) (1 + \frac{d}{d x} [- 1]) + (x - 1) \frac{d}{d x} [x + 1]$

Step 4.3.5.1.3.3

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

$(x + 1) (1 + 0) + (x - 1) \frac{d}{d x} [x + 1]$

Step 4.3.5.1.3.4

Simplify the expression.

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Step 4.3.5.1.3.4.1

Add $1$ and $0$ .

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

Step 4.3.5.1.3.4.2

Multiply $x + 1$ by $1$ .

$x + 1 + (x - 1) \frac{d}{d x} [x + 1]$

Step 4.3.5.1.3.5

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

$x + 1 + (x - 1) (\frac{d}{d x} [x] + \frac{d}{d x} [1])$

Step 4.3.5.1.3.6

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

$x + 1 + (x - 1) (1 + \frac{d}{d x} [1])$

Step 4.3.5.1.3.7

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

$x + 1 + (x - 1) (1 + 0)$

Step 4.3.5.1.3.8

Simplify by adding terms.

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Step 4.3.5.1.3.8.1

Add $1$ and $0$ .

$x + 1 + (x - 1) \cdot 1$

Step 4.3.5.1.3.8.2

Multiply $x - 1$ by $1$ .

$x + 1 + x - 1$

Step 4.3.5.1.3.8.3

Add $x$ and $x$ .

$2 x + 1 - 1$

Step 4.3.5.1.3.8.4

Simplify by subtracting numbers.

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Step 4.3.5.1.3.8.4.1

Subtract $1$ from $1$ .

$2 x + 0$

Step 4.3.5.1.3.8.4.2

Add $2 x$ and $0$ .

$2 x$

Step 4.3.5.2

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

$y + C_{1} = - 2 \int \frac{1}{u_{1}} \cdot \frac{1}{2} d u_{1} + \int - \frac{3}{(x + 1) (x - 1)} d x$

Step 4.3.6

Simplify.

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Step 4.3.6.1

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

$y + C_{1} = - 2 \int \frac{1}{u_{1} \cdot 2} d u_{1} + \int - \frac{3}{(x + 1) (x - 1)} d x$

Step 4.3.6.2

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

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

Step 4.3.7

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

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

Step 4.3.8

Simplify.

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Step 4.3.8.1

Combine $\frac{1}{2}$ and $- 2$ .

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

Step 4.3.8.2

Cancel the common factor of $- 2$ and $2$ .

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Step 4.3.8.2.1

Factor $2$ out of $- 2$ .

$y + C_{1} = \frac{2 \cdot - 1}{2} \int \frac{1}{u_{1}} d u_{1} + \int - \frac{3}{(x + 1) (x - 1)} d x$

Step 4.3.8.2.2

Cancel the common factors.

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Step 4.3.8.2.2.1

Factor $2$ out of $2$ .

$y + C_{1} = \frac{2 \cdot - 1}{2 (1)} \int \frac{1}{u_{1}} d u_{1} + \int - \frac{3}{(x + 1) (x - 1)} d x$

Step 4.3.8.2.2.2

Cancel the common factor.

$y + C_{1} = \frac{2 \cdot - 1}{2 \cdot 1} \int \frac{1}{u_{1}} d u_{1} + \int - \frac{3}{(x + 1) (x - 1)} d x$

Step 4.3.8.2.2.3

Rewrite the expression.

$y + C_{1} = \frac{- 1}{1} \int \frac{1}{u_{1}} d u_{1} + \int - \frac{3}{(x + 1) (x - 1)} d x$

Step 4.3.8.2.2.4

Divide $- 1$ by $1$ .

$y + C_{1} = - \int \frac{1}{u_{1}} d u_{1} + \int - \frac{3}{(x + 1) (x - 1)} d x$

Step 4.3.9

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

$y + C_{1} = - (\ln (| u_{1} |) + C_{2}) + \int - \frac{3}{(x + 1) (x - 1)} d x$

Step 4.3.10

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

$y + C_{1} = - (\ln (| u_{1} |) + C_{2}) - \int \frac{3}{(x + 1) (x - 1)} d x$

Step 4.3.11

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

$y + C_{1} = - (\ln (| u_{1} |) + C_{2}) - (3 \int \frac{1}{(x + 1) (x - 1)} d x)$

Step 4.3.12

Multiply $3$ by $- 1$ .

$y + C_{1} = - (\ln (| u_{1} |) + C_{2}) - 3 \int \frac{1}{(x + 1) (x - 1)} d x$

Step 4.3.13

Write the fraction using partial fraction decomposition.

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Step 4.3.13.1

Decompose the fraction and multiply through by the common denominator.

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Step 4.3.13.1.1

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 $A$ .

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

Step 4.3.13.1.2

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

Step 4.3.13.1.3

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

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

Step 4.3.13.1.4

Cancel the common factor of $x + 1$ .

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Step 4.3.13.1.4.1

Cancel the common factor.

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

Step 4.3.13.1.4.2

Rewrite the expression.

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

Step 4.3.13.1.5

Cancel the common factor of $x - 1$ .

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Step 4.3.13.1.5.1

Cancel the common factor.

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

Step 4.3.13.1.5.2

Rewrite the expression.

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

Step 4.3.13.1.6

Simplify each term.

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Step 4.3.13.1.6.1

Cancel the common factor of $x + 1$ .

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Step 4.3.13.1.6.1.1

Cancel the common factor.

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

Step 4.3.13.1.6.1.2

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

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

Step 4.3.13.1.6.2

Apply the distributive property.

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

Step 4.3.13.1.6.3

Move $- 1$ to the left of $A$ .

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

Step 4.3.13.1.6.4

Rewrite $- 1 A$ as $- A$ .

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

Step 4.3.13.1.6.5

Cancel the common factor of $x - 1$ .

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Step 4.3.13.1.6.5.1

Cancel the common factor.

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

Step 4.3.13.1.6.5.2

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

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

Step 4.3.13.1.6.6

Apply the distributive property.

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

Step 4.3.13.1.6.7

Multiply $B$ by $1$ .

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

Step 4.3.13.1.7

Move $- A$ .

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

Step 4.3.13.2

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

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Step 4.3.13.2.1

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.

$0 = A + B$

Step 4.3.13.2.2

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.

$1 = - 1 A + B$

Step 4.3.13.2.3

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

$0 = A + B$

$1 = - 1 A + B$

$0 = A + B$

$1 = - 1 A + B$

Step 4.3.13.3

Solve the system of equations.

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Step 4.3.13.3.1

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

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Step 4.3.13.3.1.1

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

$A + B = 0$

$1 = - 1 A + B$

Step 4.3.13.3.1.2

Subtract $B$ from both sides of the equation.

$A = - B$

$1 = - 1 A + B$

$A = - B$

$1 = - 1 A + B$

Step 4.3.13.3.2

Replace all occurrences of $A$ with $- B$ in each equation.

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Step 4.3.13.3.2.1

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

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

$A = - B$

Step 4.3.13.3.2.2

Simplify the right side.

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Step 4.3.13.3.2.2.1

Simplify $- 1 (- B) + B$ .

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Step 4.3.13.3.2.2.1.1

Multiply $- 1 (- B)$ .

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Step 4.3.13.3.2.2.1.1.1

Multiply $- 1$ by $- 1$ .

$1 = 1 B + B$

$A = - B$

Step 4.3.13.3.2.2.1.1.2

Multiply $B$ by $1$ .

$1 = B + B$

$A = - B$

$1 = B + B$

$A = - B$

Step 4.3.13.3.2.2.1.2

Add $B$ and $B$ .

$1 = 2 B$

$A = - B$

$1 = 2 B$

$A = - B$

$1 = 2 B$

$A = - B$

$1 = 2 B$

$A = - B$

Step 4.3.13.3.3

Solve for $B$ in $1 = 2 B$ .

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Step 4.3.13.3.3.1

Rewrite the equation as $2 B = 1$ .

$2 B = 1$

$A = - B$

Step 4.3.13.3.3.2

Divide each term in $2 B = 1$ by $2$ and simplify.

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Step 4.3.13.3.3.2.1

Divide each term in $2 B = 1$ by $2$ .

$\frac{2 B}{2} = \frac{1}{2}$

$A = - B$

Step 4.3.13.3.3.2.2

Simplify the left side.

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Step 4.3.13.3.3.2.2.1

Cancel the common factor of $2$ .

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Step 4.3.13.3.3.2.2.1.1

Cancel the common factor.

$\frac{2 B}{2} = \frac{1}{2}$

$A = - B$

Step 4.3.13.3.3.2.2.1.2

Divide $B$ by $1$ .

$B = \frac{1}{2}$

$A = - B$

$B = \frac{1}{2}$

$A = - B$

$B = \frac{1}{2}$

$A = - B$

$B = \frac{1}{2}$

$A = - B$

$B = \frac{1}{2}$

$A = - B$

Step 4.3.13.3.4

Replace all occurrences of $B$ with $\frac{1}{2}$ in each equation.

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Step 4.3.13.3.4.1

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

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

$B = \frac{1}{2}$

Step 4.3.13.3.4.2

Simplify the right side.

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Step 4.3.13.3.4.2.1

Multiply $- 1$ by $\frac{1}{2}$ .

$A = - \frac{1}{2}$

$B = \frac{1}{2}$

$A = - \frac{1}{2}$

$B = \frac{1}{2}$

$A = - \frac{1}{2}$

$B = \frac{1}{2}$

Step 4.3.13.3.5

List all of the solutions.

$A = - \frac{1}{2}, B = \frac{1}{2}$

Step 4.3.13.4

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

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

Step 4.3.13.5

Simplify.

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Step 4.3.13.5.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 4.3.13.5.2

Multiply $\frac{1}{x + 1}$ by $\frac{1}{2}$ .

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

Step 4.3.13.5.3

Move $2$ to the left of $x + 1$ .

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

Step 4.3.13.5.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 4.3.13.5.5

Multiply $\frac{1}{2}$ by $\frac{1}{x - 1}$ .

$y + C_{1} = - (\ln (| u_{1} |) + C_{2}) - 3 \int - \frac{1}{2 (x + 1)} + \frac{1}{2 (x - 1)} d x$

Step 4.3.14

Split the single integral into multiple integrals.

$y + C_{1} = - (\ln (| u_{1} |) + C_{2}) - 3 (\int - \frac{1}{2 (x + 1)} d x + \int \frac{1}{2 (x - 1)} d x)$

Step 4.3.15

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

$y + C_{1} = - (\ln (| u_{1} |) + C_{2}) - 3 (- \int \frac{1}{2 (x + 1)} d x + \int \frac{1}{2 (x - 1)} d x)$

Step 4.3.16

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

$y + C_{1} = - (\ln (| u_{1} |) + C_{2}) - 3 (- (\frac{1}{2} \int \frac{1}{x + 1} d x) + \int \frac{1}{2 (x - 1)} d x)$

Step 4.3.17

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

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Step 4.3.17.1

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

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Step 4.3.17.1.1

Differentiate $x + 1$ .

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

Step 4.3.17.1.2

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 4.3.17.1.3

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 4.3.17.1.4

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

$1 + 0$

Step 4.3.17.1.5

Add $1$ and $0$ .

$1$

Step 4.3.17.2

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

$y + C_{1} = - (\ln (| u_{1} |) + C_{2}) - 3 (- \frac{1}{2} \int \frac{1}{u_{2}} d u_{2} + \int \frac{1}{2 (x - 1)} d x)$

Step 4.3.18

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

$y + C_{1} = - (\ln (| u_{1} |) + C_{2}) - 3 (- \frac{1}{2} (\ln (| u_{2} |) + C_{3}) + \int \frac{1}{2 (x - 1)} d x)$

Step 4.3.19

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

$y + C_{1} = - (\ln (| u_{1} |) + C_{2}) - 3 (- \frac{1}{2} (\ln (| u_{2} |) + C_{3}) + \frac{1}{2} \int \frac{1}{x - 1} d x)$

Step 4.3.20

Let $u_{3} = x - 1$ . Then $d u_{3} = d x$ . Rewrite using $u_{3}$ and $d$ $u_{3}$ .

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Step 4.3.20.1

Let $u_{3} = x - 1$ . Find $\frac{d u_{3}}{d x}$ .

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Step 4.3.20.1.1

Differentiate $x - 1$ .

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

Step 4.3.20.1.2

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 4.3.20.1.3

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 4.3.20.1.4

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

$1 + 0$

Step 4.3.20.1.5

Add $1$ and $0$ .

$1$

Step 4.3.20.2

Rewrite the problem using $u_{3}$ and $d u_{3}$ .

$y + C_{1} = - (\ln (| u_{1} |) + C_{2}) - 3 (- \frac{1}{2} (\ln (| u_{2} |) + C_{3}) + \frac{1}{2} \int \frac{1}{u_{3}} d u_{3})$

Step 4.3.21

The integral of $\frac{1}{u_{3}}$ with respect to $u_{3}$ is $\ln (| u_{3} |)$ .

$y + C_{1} = - (\ln (| u_{1} |) + C_{2}) - 3 (- \frac{1}{2} (\ln (| u_{2} |) + C_{3}) + \frac{1}{2} (\ln (| u_{3} |) + C_{4}))$

Step 4.3.22

Simplify.

$y + C_{1} = - \ln (| u_{1} |) - 3 (- \frac{1}{2} \ln (| u_{2} |) + \frac{1}{2} \ln (| u_{3} |)) + C_{5}$

Step 4.3.23

Substitute back in for each integration substitution variable.

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Step 4.3.23.1

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

$y + C_{1} = - \ln (| (x + 1) (x - 1) |) - 3 (- \frac{1}{2} \ln (| u_{2} |) + \frac{1}{2} \ln (| u_{3} |)) + C_{5}$

Step 4.3.23.2

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

$y + C_{1} = - \ln (| (x + 1) (x - 1) |) - 3 (- \frac{1}{2} \ln (| x + 1 |) + \frac{1}{2} \ln (| u_{3} |)) + C_{5}$

Step 4.3.23.3

Replace all occurrences of $u_{3}$ with $x - 1$ .

$y + C_{1} = - \ln (| (x + 1) (x - 1) |) - 3 (- \frac{1}{2} \ln (| x + 1 |) + \frac{1}{2} \ln (| x - 1 |)) + C_{5}$

Step 4.4

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

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