Calculus Examples

$\frac{d y}{d x} + \frac{4}{x^{2} - 1} y = 0$

Step 1

The integrating factor is defined by the formula $e^{\int P (x) d x}$ , where $P (x) = \frac{4}{x^{2} - 1}$ .

Tap for more steps...

Step 1.1

Set up the integration.

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

Step 1.2

Integrate $\frac{4}{x^{2} - 1}$ .

Tap for more steps...

Step 1.2.1

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

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

Step 1.2.2

Write the fraction using partial fraction decomposition.

Tap for more steps...

Step 1.2.2.1

Decompose the fraction and multiply through by the common denominator.

Tap for more steps...

Step 1.2.2.1.1

Factor the fraction.

Tap for more steps...

Step 1.2.2.1.1.1

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

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

Step 1.2.2.1.1.2

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 $b = 1$ .

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

Step 1.2.2.1.2

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 $A$ .

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

Step 1.2.2.1.3

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

Step 1.2.2.1.4

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 1.2.2.1.5

Cancel the common factor of $x + 1$ .

Tap for more steps...

Step 1.2.2.1.5.1

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 1.2.2.1.5.2

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 1.2.2.1.6

Cancel the common factor of $x - 1$ .

Tap for more steps...

Step 1.2.2.1.6.1

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 1.2.2.1.6.2

Rewrite the expression.

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

Step 1.2.2.1.7

Simplify each term.

Tap for more steps...

Step 1.2.2.1.7.1

Cancel the common factor of $x + 1$ .

Tap for more steps...

Step 1.2.2.1.7.1.1

Cancel the common factor.

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

Step 1.2.2.1.7.1.2

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

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

Step 1.2.2.1.7.2

Apply the distributive property.

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

Step 1.2.2.1.7.3

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

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

Step 1.2.2.1.7.4

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

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

Step 1.2.2.1.7.5

Cancel the common factor of $x - 1$ .

Tap for more steps...

Step 1.2.2.1.7.5.1

Cancel the common factor.

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

Step 1.2.2.1.7.5.2

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

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

Step 1.2.2.1.7.6

Apply the distributive property.

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

Step 1.2.2.1.7.7

Multiply $B$ by $1$ .

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

Step 1.2.2.1.8

Move $- A$ .

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

Step 1.2.2.2

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

Tap for more steps...

Step 1.2.2.2.1

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 1.2.2.2.2

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 1.2.2.2.3

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 1.2.2.3

Solve the system of equations.

Tap for more steps...

Step 1.2.2.3.1

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

Tap for more steps...

Step 1.2.2.3.1.1

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

$A + B = 0$

$1 = - 1 A + B$

Step 1.2.2.3.1.2

Subtract $B$ from both sides of the equation.

$A = - B$

$1 = - 1 A + B$

$A = - B$

$1 = - 1 A + B$

Step 1.2.2.3.2

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

Tap for more steps...

Step 1.2.2.3.2.1

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

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

$A = - B$

Step 1.2.2.3.2.2

Simplify the right side.

Tap for more steps...

Step 1.2.2.3.2.2.1

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

Tap for more steps...

Step 1.2.2.3.2.2.1.1

Multiply $- 1 (- B)$ .

Tap for more steps...

Step 1.2.2.3.2.2.1.1.1

Multiply $- 1$ by $- 1$ .

$1 = 1 B + B$

$A = - B$

Step 1.2.2.3.2.2.1.1.2

Multiply $B$ by $1$ .

$1 = B + B$

$A = - B$

$1 = B + B$

$A = - B$

Step 1.2.2.3.2.2.1.2

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 1.2.2.3.3

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

Tap for more steps...

Step 1.2.2.3.3.1

Rewrite the equation as $2 B = 1$ .

$2 B = 1$

$A = - B$

Step 1.2.2.3.3.2

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

Tap for more steps...

Step 1.2.2.3.3.2.1

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

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

$A = - B$

Step 1.2.2.3.3.2.2

Simplify the left side.

Tap for more steps...

Step 1.2.2.3.3.2.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 1.2.2.3.3.2.2.1.1

Cancel the common factor.

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

$A = - B$

Step 1.2.2.3.3.2.2.1.2

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 1.2.2.3.4

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

Tap for more steps...

Step 1.2.2.3.4.1

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

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

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

Step 1.2.2.3.4.2

Simplify the right side.

Tap for more steps...

Step 1.2.2.3.4.2.1

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 1.2.2.3.5

List all of the solutions.

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

Step 1.2.2.4

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 1.2.2.5

Simplify.

Tap for more steps...

Step 1.2.2.5.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.2.2.5.2

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

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

Step 1.2.2.5.3

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

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

Step 1.2.2.5.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.2.2.5.5

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

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

Step 1.2.3

Split the single integral into multiple integrals.

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

Step 1.2.4

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

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

Step 1.2.5

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

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

Step 1.2.6

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

Tap for more steps...

Step 1.2.6.1

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

Tap for more steps...

Step 1.2.6.1.1

Differentiate $x + 1$ .

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

Step 1.2.6.1.2

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 1.2.6.1.3

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 1.2.6.1.4

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

$1 + 0$

Step 1.2.6.1.5

Add $1$ and $0$ .

$1$

Step 1.2.6.2

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

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

Step 1.2.7

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

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

Step 1.2.8

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

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

Step 1.2.9

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

Tap for more steps...

Step 1.2.9.1

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

Tap for more steps...

Step 1.2.9.1.1

Differentiate $x - 1$ .

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

Step 1.2.9.1.2

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 1.2.9.1.3

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 1.2.9.1.4

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

$1 + 0$

Step 1.2.9.1.5

Add $1$ and $0$ .

$1$

Step 1.2.9.2

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

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

Step 1.2.10

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

$e^{4 (- \frac{1}{2} (\ln (| u_{1} |) + C) + \frac{1}{2} (\ln (| u_{2} |) + C))}$

Step 1.2.11

Simplify.

$e^{4 (- \frac{1}{2} \ln (| u_{1} |) + \frac{1}{2} \ln (| u_{2} |)) + C}$

Step 1.2.12

Substitute back in for each integration substitution variable.

Tap for more steps...

Step 1.2.12.1

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

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

Step 1.2.12.2

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

$e^{4 (- \frac{1}{2} \ln (| x + 1 |) + \frac{1}{2} \ln (| x - 1 |)) + C}$

Step 1.2.13

Simplify.

Tap for more steps...

Step 1.2.13.1

Simplify each term.

Tap for more steps...

Step 1.2.13.1.1

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

$e^{4 (- \frac{\ln (| x + 1 |)}{2} + \frac{1}{2} \ln (| x - 1 |)) + C}$

Step 1.2.13.1.2

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

$e^{4 (- \frac{\ln (| x + 1 |)}{2} + \frac{\ln (| x - 1 |)}{2}) + C}$

Step 1.2.13.2

Combine the numerators over the common denominator.

$e^{4 \frac{- \ln (| x + 1 |) + \ln (| x - 1 |)}{2} + C}$

Step 1.2.13.3

Cancel the common factor of $2$ .

Tap for more steps...

Step 1.2.13.3.1

Factor $2$ out of $4$ .

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

Step 1.2.13.3.2

Cancel the common factor.

$e^{2 \cdot 2 \frac{- \ln (| x + 1 |) + \ln (| x - 1 |)}{2} + C}$

Step 1.2.13.3.3

Rewrite the expression.

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

Step 1.2.13.4

Apply the distributive property.

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

Step 1.2.13.5

Multiply $- 1$ by $2$ .

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

Step 1.3

Remove the constant of integration.

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

Step 1.4

Use the logarithmic power rule.

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

Step 1.5

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

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

Step 1.6

Exponentiation and log are inverse functions.

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

Step 1.7

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 1.8

Rewrite ${(x - 1)}^{2}$ as $(x - 1) (x - 1)$ .

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

Step 1.9

Expand $(x - 1) (x - 1)$ using the FOIL Method.

Tap for more steps...

Step 1.9.1

Apply the distributive property.

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

Step 1.9.2

Apply the distributive property.

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

Step 1.9.3

Apply the distributive property.

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

Step 1.10

Simplify and combine like terms.

Tap for more steps...

Step 1.10.1

Simplify each term.

Tap for more steps...

Step 1.10.1.1

Multiply $x$ by $x$ .

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

Step 1.10.1.2

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

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

Step 1.10.1.3

Rewrite $- 1 x$ as $- x$ .

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

Step 1.10.1.4

Rewrite $- 1 x$ as $- x$ .

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

Step 1.10.1.5

Multiply $- 1$ by $- 1$ .

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

Step 1.10.2

Subtract $x$ from $- x$ .

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

Step 1.11

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

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

Step 1.12

Factor using the perfect square rule.

Tap for more steps...

Step 1.12.1

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

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

Step 1.12.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Step 1.12.3

Rewrite the polynomial.

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

Step 1.12.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

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

Step 2

Multiply each term by the integrating factor $\frac{{(x - 1)}^{2}}{{(x + 1)}^{2}}$ .

Tap for more steps...

Step 2.1

Multiply each term by $\frac{{(x - 1)}^{2}}{{(x + 1)}^{2}}$ .

$\frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \frac{d y}{d x} + \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} (\frac{4}{x^{2} - 1} y) = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.2

Simplify each term.

Tap for more steps...

Step 2.2.1

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

$\frac{{(x - 1)}^{2} \frac{d y}{d x}}{{(x + 1)}^{2}} + \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} (\frac{4}{x^{2} - 1} y) = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.2.2

Simplify the denominator.

Tap for more steps...

Step 2.2.2.1

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

$\frac{{(x - 1)}^{2} \frac{d y}{d x}}{{(x + 1)}^{2}} + \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} (\frac{4}{x^{2} - 1^{2}} y) = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.2.2.2

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 $b = 1$ .

$\frac{{(x - 1)}^{2} \frac{d y}{d x}}{{(x + 1)}^{2}} + \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} (\frac{4}{(x + 1) (x - 1)} y) = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.2.3

Combine $\frac{4}{(x + 1) (x - 1)}$ and $y$ .

$\frac{{(x - 1)}^{2} \frac{d y}{d x}}{{(x + 1)}^{2}} + \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot \frac{4 y}{(x + 1) (x - 1)} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.2.4

Combine.

$\frac{{(x - 1)}^{2} \frac{d y}{d x}}{{(x + 1)}^{2}} + \frac{{(x - 1)}^{2} (4 y)}{{(x + 1)}^{2} ((x + 1) (x - 1))} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.2.5

Multiply ${(x + 1)}^{2}$ by $x + 1$ by adding the exponents.

Tap for more steps...

Step 2.2.5.1

Move $x + 1$ .

$\frac{{(x - 1)}^{2} \frac{d y}{d x}}{{(x + 1)}^{2}} + \frac{{(x - 1)}^{2} (4 y)}{(x + 1) {(x + 1)}^{2} (x - 1)} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.2.5.2

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

Tap for more steps...

Step 2.2.5.2.1

Raise $x + 1$ to the power of $1$ .

$\frac{{(x - 1)}^{2} \frac{d y}{d x}}{{(x + 1)}^{2}} + \frac{{(x - 1)}^{2} (4 y)}{{(x + 1)}^{1} {(x + 1)}^{2} (x - 1)} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.2.5.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{{(x - 1)}^{2} \frac{d y}{d x}}{{(x + 1)}^{2}} + \frac{{(x - 1)}^{2} (4 y)}{{(x + 1)}^{1 + 2} (x - 1)} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.2.5.3

Add $1$ and $2$ .

$\frac{{(x - 1)}^{2} \frac{d y}{d x}}{{(x + 1)}^{2}} + \frac{{(x - 1)}^{2} (4 y)}{{(x + 1)}^{3} (x - 1)} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.2.6

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

Tap for more steps...

Step 2.2.6.1

Factor $x - 1$ out of ${(x - 1)}^{2} (4 y)$ .

$\frac{{(x - 1)}^{2} \frac{d y}{d x}}{{(x + 1)}^{2}} + \frac{(x - 1) ((x - 1) (4 y))}{{(x + 1)}^{3} (x - 1)} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.2.6.2

Cancel the common factors.

Tap for more steps...

Step 2.2.6.2.1

Factor $x - 1$ out of ${(x + 1)}^{3} (x - 1)$ .

$\frac{{(x - 1)}^{2} \frac{d y}{d x}}{{(x + 1)}^{2}} + \frac{(x - 1) ((x - 1) (4 y))}{(x - 1) {(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.2.6.2.2

Cancel the common factor.

$\frac{{(x - 1)}^{2} \frac{d y}{d x}}{{(x + 1)}^{2}} + \frac{(x - 1) ((x - 1) (4 y))}{(x - 1) {(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.2.6.2.3

Rewrite the expression.

$\frac{{(x - 1)}^{2} \frac{d y}{d x}}{{(x + 1)}^{2}} + \frac{(x - 1) (4 y)}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.2.7

Move $4$ to the left of $x - 1$ .

$\frac{{(x - 1)}^{2} \frac{d y}{d x}}{{(x + 1)}^{2}} + \frac{4 (x - 1) y}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.3

To write $\frac{{(x - 1)}^{2} \frac{d y}{d x}}{{(x + 1)}^{2}}$ as a fraction with a common denominator, multiply by $\frac{x + 1}{x + 1}$ .

$\frac{{(x - 1)}^{2} \frac{d y}{d x}}{{(x + 1)}^{2}} \cdot \frac{x + 1}{x + 1} + \frac{4 (x - 1) y}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.4

Write each expression with a common denominator of ${(x + 1)}^{3}$ , by multiplying each by an appropriate factor of $1$ .

Tap for more steps...

Step 2.4.1

Multiply $\frac{{(x - 1)}^{2} \frac{d y}{d x}}{{(x + 1)}^{2}}$ by $\frac{x + 1}{x + 1}$ .

$\frac{{(x - 1)}^{2} \frac{d y}{d x} (x + 1)}{{(x + 1)}^{2} (x + 1)} + \frac{4 (x - 1) y}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.4.2

Multiply ${(x + 1)}^{2}$ by $x + 1$ by adding the exponents.

Tap for more steps...

Step 2.4.2.1

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

Tap for more steps...

Step 2.4.2.1.1

Raise $x + 1$ to the power of $1$ .

$\frac{{(x - 1)}^{2} \frac{d y}{d x} (x + 1)}{{(x + 1)}^{2} {(x + 1)}^{1}} + \frac{4 (x - 1) y}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.4.2.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{{(x - 1)}^{2} \frac{d y}{d x} (x + 1)}{{(x + 1)}^{2 + 1}} + \frac{4 (x - 1) y}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.4.2.2

Add $2$ and $1$ .

$\frac{{(x - 1)}^{2} \frac{d y}{d x} (x + 1)}{{(x + 1)}^{3}} + \frac{4 (x - 1) y}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.5

Combine the numerators over the common denominator.

$\frac{{(x - 1)}^{2} \frac{d y}{d x} (x + 1) + 4 (x - 1) y}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.6

Simplify the numerator.

Tap for more steps...

Step 2.6.1

Factor $x - 1$ out of ${(x - 1)}^{2} \frac{d y}{d x} (x + 1) + 4 (x - 1) y$ .

Tap for more steps...

Step 2.6.1.1

Factor $x - 1$ out of ${(x - 1)}^{2} \frac{d y}{d x} (x + 1)$ .

$\frac{(x - 1) ((x - 1) \frac{d y}{d x} (x + 1)) + 4 (x - 1) y}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.6.1.2

Factor $x - 1$ out of $4 (x - 1) y$ .

$\frac{(x - 1) ((x - 1) \frac{d y}{d x} (x + 1)) + (x - 1) (4 y)}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.6.1.3

Factor $x - 1$ out of $(x - 1) ((x - 1) \frac{d y}{d x} (x + 1)) + (x - 1) (4 y)$ .

$\frac{(x - 1) ((x - 1) \frac{d y}{d x} (x + 1) + 4 y)}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.6.2

Apply the distributive property.

$\frac{(x - 1) ((x \frac{d y}{d x} - 1 \frac{d y}{d x}) (x + 1) + 4 y)}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.6.3

Rewrite $- 1 \frac{d y}{d x}$ as $- \frac{d y}{d x}$ .

$\frac{(x - 1) ((x \frac{d y}{d x} - \frac{d y}{d x}) (x + 1) + 4 y)}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.6.4

Expand $(x \frac{d y}{d x} - \frac{d y}{d x}) (x + 1)$ using the FOIL Method.

Tap for more steps...

Step 2.6.4.1

Apply the distributive property.

$\frac{(x - 1) (x \frac{d y}{d x} (x + 1) - \frac{d y}{d x} (x + 1) + 4 y)}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.6.4.2

Apply the distributive property.

$\frac{(x - 1) (x \frac{d y}{d x} x + x \frac{d y}{d x} \cdot 1 - \frac{d y}{d x} (x + 1) + 4 y)}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.6.4.3

Apply the distributive property.

$\frac{(x - 1) (x \frac{d y}{d x} x + x \frac{d y}{d x} \cdot 1 - \frac{d y}{d x} x - \frac{d y}{d x} \cdot 1 + 4 y)}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.6.5

Simplify and combine like terms.

Tap for more steps...

Step 2.6.5.1

Simplify each term.

Tap for more steps...

Step 2.6.5.1.1

Multiply $x$ by $x$ by adding the exponents.

Tap for more steps...

Step 2.6.5.1.1.1

Move $x$ .

$\frac{(x - 1) (x \cdot x \frac{d y}{d x} + x \frac{d y}{d x} \cdot 1 - \frac{d y}{d x} x - \frac{d y}{d x} \cdot 1 + 4 y)}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.6.5.1.1.2

Multiply $x$ by $x$ .

$\frac{(x - 1) (x^{2} \frac{d y}{d x} + x \frac{d y}{d x} \cdot 1 - \frac{d y}{d x} x - \frac{d y}{d x} \cdot 1 + 4 y)}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.6.5.1.2

Multiply $x$ by $1$ .

$\frac{(x - 1) (x^{2} \frac{d y}{d x} + x \frac{d y}{d x} - \frac{d y}{d x} x - \frac{d y}{d x} \cdot 1 + 4 y)}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.6.5.1.3

Multiply $- 1$ by $1$ .

$\frac{(x - 1) (x^{2} \frac{d y}{d x} + x \frac{d y}{d x} - \frac{d y}{d x} x - \frac{d y}{d x} + 4 y)}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.6.5.2

Subtract $\frac{d y}{d x} x$ from $x \frac{d y}{d x}$ .

Tap for more steps...

Step 2.6.5.2.1

Move $\frac{d y}{d x}$ .

$\frac{(x - 1) (x^{2} \frac{d y}{d x} + x \frac{d y}{d x} - 1 x \frac{d y}{d x} - \frac{d y}{d x} + 4 y)}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.6.5.2.2

Subtract $x \frac{d y}{d x}$ from $x \frac{d y}{d x}$ .

$\frac{(x - 1) (x^{2} \frac{d y}{d x} + 0 - \frac{d y}{d x} + 4 y)}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.6.5.3

Add $x^{2} \frac{d y}{d x}$ and $0$ .

$\frac{(x - 1) (x^{2} \frac{d y}{d x} - \frac{d y}{d x} + 4 y)}{{(x + 1)}^{3}} = \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} \cdot 0$

Step 2.7

Multiply $\frac{{(x - 1)}^{2}}{{(x + 1)}^{2}}$ by $0$ .

$\frac{(x - 1) (x^{2} \frac{d y}{d x} - \frac{d y}{d x} + 4 y)}{{(x + 1)}^{3}} = 0$

Step 3

Rewrite the left side as a result of differentiating a product.

$\frac{d}{d x} [\frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} y] = 0$

Step 4

Set up an integral on each side.

$\int \frac{d}{d x} [\frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} y] d x = \int 0 d x$

Step 5

Integrate the left side.

$\frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} y = \int 0 d x$

Step 6

Integrate the right side.

Tap for more steps...

Step 6.1

The integral of $0$ with respect to $x$ is $0$ .

$\frac{{(x - 1)}^{2}}{{(x + 1)}^{2}} y = 0 + C$

Step 6.2

Add $0$ and $C$ .

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

Step 7

Solve for $y$ .

Tap for more steps...

Step 7.1

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

Tap for more steps...

Step 7.1.1

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

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

Step 7.1.2

Reorder factors in $\frac{{(x - 1)}^{2} y}{{(x + 1)}^{2}}$ .

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

Step 7.2

Multiply both sides by ${(x + 1)}^{2}$ .

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

Step 7.3

Simplify the left side.

Tap for more steps...

Step 7.3.1

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

Tap for more steps...

Step 7.3.1.1

Cancel the common factor.

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

Step 7.3.1.2

Rewrite the expression.

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

Step 7.4

Divide each term in $y {(x - 1)}^{2} = C {(x + 1)}^{2}$ by ${(x - 1)}^{2}$ and simplify.

Tap for more steps...

Step 7.4.1

Divide each term in $y {(x - 1)}^{2} = C {(x + 1)}^{2}$ by ${(x - 1)}^{2}$ .

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

Step 7.4.2

Simplify the left side.

Tap for more steps...

Step 7.4.2.1

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

Tap for more steps...

Step 7.4.2.1.1

Cancel the common factor.

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

Step 7.4.2.1.2

Divide $y$ by $1$ .

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