Calculus Examples

$\frac{d y}{d x} = \frac{5}{{(x + 2)}^{2} e^{y - 1}}$ , $y (3) = 1$

Step 1

Separate the variables.

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Step 1.1

Regroup factors.

$\frac{d y}{d x} = \frac{5}{{(x + 2)}^{2}} \cdot \frac{1}{e^{y - 1}}$

Step 1.2

Multiply both sides by $e^{y - 1}$ .

$e^{y - 1} \frac{d y}{d x} = e^{y - 1} (\frac{5}{{(x + 2)}^{2}} \cdot \frac{1}{e^{y - 1}})$

Step 1.3

Simplify.

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Step 1.3.1

Combine.

$e^{y - 1} \frac{d y}{d x} = e^{y - 1} \frac{5 \cdot 1}{{(x + 2)}^{2} e^{y - 1}}$

Step 1.3.2

Cancel the common factor of $e^{y - 1}$ .

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Step 1.3.2.1

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

$e^{y - 1} \frac{d y}{d x} = e^{y - 1} \frac{5 \cdot 1}{e^{y - 1} {(x + 2)}^{2}}$

Step 1.3.2.2

Cancel the common factor.

$e^{y - 1} \frac{d y}{d x} = e^{y - 1} \frac{5 \cdot 1}{e^{y - 1} {(x + 2)}^{2}}$

Step 1.3.2.3

Rewrite the expression.

$e^{y - 1} \frac{d y}{d x} = \frac{5 \cdot 1}{{(x + 2)}^{2}}$

Step 1.3.3

Multiply $5$ by $1$ .

$e^{y - 1} \frac{d y}{d x} = \frac{5}{{(x + 2)}^{2}}$

Step 1.4

Rewrite the equation.

$e^{y - 1} d y = \frac{5}{{(x + 2)}^{2}} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int e^{y - 1} d y = \int \frac{5}{{(x + 2)}^{2}} d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Differentiate $y - 1$ .

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

Step 2.2.1.1.2

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

$\frac{d}{d y} [y] + \frac{d}{d y} [- 1]$

Step 2.2.1.1.3

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

$1 + \frac{d}{d y} [- 1]$

Step 2.2.1.1.4

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

$1 + 0$

Step 2.2.1.1.5

Add $1$ and $0$ .

$1$

Step 2.2.1.2

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

$\int e^{u_{1}} d u_{1} = \int \frac{5}{{(x + 2)}^{2}} d x$

Step 2.2.2

The integral of $e^{u_{1}}$ with respect to $u_{1}$ is $e^{u_{1}}$ .

$e^{u_{1}} + C_{1} = \int \frac{5}{{(x + 2)}^{2}} d x$

Step 2.2.3

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

$e^{y - 1} + C_{1} = \int \frac{5}{{(x + 2)}^{2}} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

$e^{y - 1} + C_{1} = 5 \int \frac{1}{{(x + 2)}^{2}} d x$

Step 2.3.2

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

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Step 2.3.2.1

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

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Step 2.3.2.1.1

Differentiate $x + 2$ .

$\frac{d}{d x} [x + 2]$

Step 2.3.2.1.2

By the Sum Rule, the derivative of $x + 2$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [2]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [2]$

Step 2.3.2.1.3

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

$1 + \frac{d}{d x} [2]$

Step 2.3.2.1.4

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$1 + 0$

Step 2.3.2.1.5

Add $1$ and $0$ .

$1$

Step 2.3.2.2

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

$e^{y - 1} + C_{1} = 5 \int \frac{1}{{u_{2}}^{2}} d u_{2}$

Step 2.3.3

Apply basic rules of exponents.

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Step 2.3.3.1

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

$e^{y - 1} + C_{1} = 5 \int {({u_{2}}^{2})}^{- 1} d u_{2}$

Step 2.3.3.2

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

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Step 2.3.3.2.1

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

$e^{y - 1} + C_{1} = 5 \int {u_{2}}^{2 \cdot - 1} d u_{2}$

Step 2.3.3.2.2

Multiply $2$ by $- 1$ .

$e^{y - 1} + C_{1} = 5 \int {u_{2}}^{- 2} d u_{2}$

Step 2.3.4

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

$e^{y - 1} + C_{1} = 5 (- {u_{2}}^{- 1} + C_{2})$

Step 2.3.5

Simplify.

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Step 2.3.5.1

Rewrite $5 (- {u_{2}}^{- 1} + C_{2})$ as $5 (- \frac{1}{u_{2}}) + C_{2}$ .

$e^{y - 1} + C_{1} = 5 (- \frac{1}{u_{2}}) + C_{2}$

Step 2.3.5.2

Simplify.

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Step 2.3.5.2.1

Multiply $- 1$ by $5$ .

$e^{y - 1} + C_{1} = - 5 \frac{1}{u_{2}} + C_{2}$

Step 2.3.5.2.2

Combine $- 5$ and $\frac{1}{u_{2}}$ .

$e^{y - 1} + C_{1} = \frac{- 5}{u_{2}} + C_{2}$

Step 2.3.5.2.3

Move the negative in front of the fraction.

$e^{y - 1} + C_{1} = - \frac{5}{u_{2}} + C_{2}$

Step 2.3.6

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

$e^{y - 1} + C_{1} = - \frac{5}{x + 2} + C_{2}$

Step 2.4

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

$e^{y - 1} = - \frac{5}{x + 2} + K$

Step 3

Solve for $y$ .

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Step 3.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{y - 1}) = \ln (\frac{- 5 + K x + 2 K}{x + 2})$

Step 3.2

Expand the left side.

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Step 3.2.1

Expand $\ln (e^{y - 1})$ by moving $y - 1$ outside the logarithm.

$(y - 1) \ln (e) = \ln (\frac{- 5 + K x + 2 K}{x + 2})$

Step 3.2.2

The natural logarithm of $e$ is $1$ .

$(y - 1) \cdot 1 = \ln (\frac{- 5 + K x + 2 K}{x + 2})$

Step 3.2.3

Multiply $y - 1$ by $1$ .

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

Step 3.3

Simplify the right side.

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Step 3.3.1

Simplify $\ln (\frac{- 5 + K x + 2 K}{x + 2})$ .

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Step 3.3.1.1

Split the fraction $\frac{- 5 + K x + 2 K}{x + 2}$ into two fractions.

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

Step 3.3.1.2

Simplify each term.

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Step 3.3.1.2.1

Split the fraction $\frac{- 5 + K x}{x + 2}$ into two fractions.

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

Step 3.3.1.2.2

Move the negative in front of the fraction.

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

Step 3.4

Add $1$ to both sides of the equation.

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

Step 4

Simplify the constant of integration.

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

Step 5

Use the initial condition to find the value of $K$ by substituting $3$ for $x$ and $1$ for $y$ in $y = \ln (- \frac{5}{x + 2} + \frac{K x}{x + 2} + \frac{K}{x + 2}) + 1$ .

$1 = \ln (- \frac{5}{3 + 2} + \frac{K \cdot 3}{3 + 2} + \frac{K}{3 + 2}) + 1$

Step 6

Solve for $K$ .

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Step 6.1

Rewrite the equation as $\ln (- \frac{5}{3 + 2} + \frac{K \cdot 3}{3 + 2} + \frac{K}{3 + 2}) + 1 = 1$ .

$\ln (- \frac{5}{3 + 2} + \frac{K \cdot 3}{3 + 2} + \frac{K}{3 + 2}) + 1 = 1$

Step 6.2

Move all terms not containing $\ln (- \frac{5}{3 + 2} + \frac{K \cdot 3}{3 + 2} + \frac{K}{3 + 2})$ to the right side of the equation.

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Step 6.2.1

Subtract $1$ from both sides of the equation.

$\ln (- \frac{5}{3 + 2} + \frac{K \cdot 3}{3 + 2} + \frac{K}{3 + 2}) = 1 - 1$

Step 6.2.2

Subtract $1$ from $1$ .

$\ln (- \frac{5}{3 + 2} + \frac{K \cdot 3}{3 + 2} + \frac{K}{3 + 2}) = 0$

Step 6.3

To solve for $K$ , rewrite the equation using properties of logarithms.

$e^{\ln (- \frac{5}{3 + 2} + \frac{K \cdot 3}{3 + 2} + \frac{K}{3 + 2})} = e^{0}$

Step 6.4

Rewrite $\ln (- \frac{5}{3 + 2} + \frac{K \cdot 3}{3 + 2} + \frac{K}{3 + 2}) = 0$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{0} = - \frac{5}{3 + 2} + \frac{K \cdot 3}{3 + 2} + \frac{K}{3 + 2}$

Step 6.5

Solve for $K$ .

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Step 6.5.1

Rewrite the equation as $- \frac{5}{3 + 2} + \frac{K \cdot 3}{3 + 2} + \frac{K}{3 + 2} = e^{0}$ .

$- \frac{5}{3 + 2} + \frac{K \cdot 3}{3 + 2} + \frac{K}{3 + 2} = e^{0}$

Step 6.5.2

Simplify $- \frac{5}{3 + 2} + \frac{K \cdot 3}{3 + 2} + \frac{K}{3 + 2}$ .

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Step 6.5.2.1

Combine the numerators over the common denominator.

$\frac{- 5 + K \cdot 3 + K}{3 + 2} = e^{0}$

Step 6.5.2.2

Move $3$ to the left of $K$ .

$\frac{- 5 + 3 K + K}{3 + 2} = e^{0}$

Step 6.5.2.3

Simplify terms.

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Step 6.5.2.3.1

Add $3 K$ and $K$ .

$\frac{- 5 + 4 K}{3 + 2} = e^{0}$

Step 6.5.2.3.2

Add $3$ and $2$ .

$\frac{- 5 + 4 K}{5} = e^{0}$

Step 6.5.2.3.3

Rewrite $- 5$ as $- 1 (5)$ .

$\frac{- 1 (5) + 4 K}{5} = e^{0}$

Step 6.5.2.3.4

Factor $- 1$ out of $4 K$ .

$\frac{- 1 (5) - (- 4 K)}{5} = e^{0}$

Step 6.5.2.3.5

Factor $- 1$ out of $- 1 (5) - (- 4 K)$ .

$\frac{- 1 (5 - 4 K)}{5} = e^{0}$

Step 6.5.2.3.6

Move the negative in front of the fraction.

$- \frac{5 - 4 K}{5} = e^{0}$

Step 6.5.3

Anything raised to $0$ is $1$ .

$- \frac{5 - 4 K}{5} = 1$

Step 6.5.4

Multiply both sides of the equation by $- 5$ .

$- 5 (- \frac{5 - 4 K}{5}) = - 5 \cdot 1$

Step 6.5.5

Simplify both sides of the equation.

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Step 6.5.5.1

Simplify the left side.

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Step 6.5.5.1.1

Simplify $- 5 (- \frac{5 - 4 K}{5})$ .

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Step 6.5.5.1.1.1

Cancel the common factor of $5$ .

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Step 6.5.5.1.1.1.1

Move the leading negative in $- \frac{5 - 4 K}{5}$ into the numerator.

$- 5 \frac{- (5 - 4 K)}{5} = - 5 \cdot 1$

Step 6.5.5.1.1.1.2

Factor $5$ out of $- 5$ .

$5 (- 1) \frac{- (5 - 4 K)}{5} = - 5 \cdot 1$

Step 6.5.5.1.1.1.3

Cancel the common factor.

$5 \cdot - 1 \frac{- (5 - 4 K)}{5} = - 5 \cdot 1$

Step 6.5.5.1.1.1.4

Rewrite the expression.

$- - (5 - 4 K) = - 5 \cdot 1$

Step 6.5.5.1.1.2

Multiply.

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Step 6.5.5.1.1.2.1

Multiply $- 1$ by $- 1$ .

$1 (5 - 4 K) = - 5 \cdot 1$

Step 6.5.5.1.1.2.2

Multiply $5 - 4 K$ by $1$ .

$5 - 4 K = - 5 \cdot 1$

Step 6.5.5.2

Simplify the right side.

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Step 6.5.5.2.1

Multiply $- 5$ by $1$ .

$5 - 4 K = - 5$

Step 6.5.6

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

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Step 6.5.6.1

Subtract $5$ from both sides of the equation.

$- 4 K = - 5 - 5$

Step 6.5.6.2

Subtract $5$ from $- 5$ .

$- 4 K = - 10$

Step 6.5.7

Divide each term in $- 4 K = - 10$ by $- 4$ and simplify.

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Step 6.5.7.1

Divide each term in $- 4 K = - 10$ by $- 4$ .

$\frac{- 4 K}{- 4} = \frac{- 10}{- 4}$

Step 6.5.7.2

Simplify the left side.

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Step 6.5.7.2.1

Cancel the common factor of $- 4$ .

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Step 6.5.7.2.1.1

Cancel the common factor.

$\frac{- 4 K}{- 4} = \frac{- 10}{- 4}$

Step 6.5.7.2.1.2

Divide $K$ by $1$ .

$K = \frac{- 10}{- 4}$

Step 6.5.7.3

Simplify the right side.

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Step 6.5.7.3.1

Cancel the common factor of $- 10$ and $- 4$ .

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Step 6.5.7.3.1.1

Factor $- 2$ out of $- 10$ .

$K = \frac{- 2 (5)}{- 4}$

Step 6.5.7.3.1.2

Cancel the common factors.

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Step 6.5.7.3.1.2.1

Factor $- 2$ out of $- 4$ .

$K = \frac{- 2 \cdot 5}{- 2 \cdot 2}$

Step 6.5.7.3.1.2.2

Cancel the common factor.

$K = \frac{- 2 \cdot 5}{- 2 \cdot 2}$

Step 6.5.7.3.1.2.3

Rewrite the expression.

$K = \frac{5}{2}$

Step 7

Substitute $\frac{5}{2}$ for $K$ in $y = \ln (- \frac{5}{x + 2} + \frac{K x}{x + 2} + \frac{K}{x + 2}) + 1$ and simplify.

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Step 7.1

Substitute $\frac{5}{2}$ for $K$ .

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

Step 7.2

Simplify each term.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

Combine $\frac{5}{2}$ and $x$ .

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

Step 7.2.1.2

Multiply the numerator by the reciprocal of the denominator.

$y = \ln (- \frac{5}{x + 2} + \frac{5 x}{2} \cdot \frac{1}{x + 2} + \frac{K}{x + 2}) + 1$

Step 7.2.1.3

Multiply $\frac{5 x}{2}$ by $\frac{1}{x + 2}$ .

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

Step 7.2.2

To write $- \frac{5}{x + 2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$y = \ln (- \frac{5}{x + 2} \cdot \frac{2}{2} + \frac{5 x}{2 (x + 2)} + \frac{K}{x + 2}) + 1$

Step 7.2.3

Write each expression with a common denominator of $(x + 2) \cdot 2$ , by multiplying each by an appropriate factor of $1$ .

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Step 7.2.3.1

Multiply $\frac{5}{x + 2}$ by $\frac{2}{2}$ .

$y = \ln (- \frac{5 \cdot 2}{(x + 2) \cdot 2} + \frac{5 x}{2 (x + 2)} + \frac{K}{x + 2}) + 1$

Step 7.2.3.2

Reorder the factors of $(x + 2) \cdot 2$ .

$y = \ln (- \frac{5 \cdot 2}{2 (x + 2)} + \frac{5 x}{2 (x + 2)} + \frac{K}{x + 2}) + 1$

Step 7.2.4

Combine the numerators over the common denominator.

$y = \ln (\frac{- 5 \cdot 2 + 5 x}{2 (x + 2)} + \frac{K}{x + 2}) + 1$

Step 7.2.5

Multiply $- 5$ by $2$ .

$y = \ln (\frac{- 10 + 5 x}{2 (x + 2)} + \frac{K}{x + 2}) + 1$

Step 7.2.6

Factor $5$ out of $- 10 + 5 x$ .

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Step 7.2.6.1

Factor $5$ out of $- 10$ .

$y = \ln (\frac{5 \cdot - 2 + 5 x}{2 (x + 2)} + \frac{K}{x + 2}) + 1$

Step 7.2.6.2

Factor $5$ out of $5 \cdot - 2 + 5 x$ .

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

Step 7.2.7

To write $\frac{K}{x + 2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$y = \ln (\frac{5 (- 2 + x)}{2 (x + 2)} + \frac{K}{x + 2} \cdot \frac{2}{2}) + 1$

Step 7.2.8

Write each expression with a common denominator of $2 (x + 2)$ , by multiplying each by an appropriate factor of $1$ .

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Step 7.2.8.1

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

$y = \ln (\frac{5 (- 2 + x)}{2 (x + 2)} + \frac{K \cdot 2}{(x + 2) \cdot 2}) + 1$

Step 7.2.8.2

Reorder the factors of $(x + 2) \cdot 2$ .

$y = \ln (\frac{5 (- 2 + x)}{2 (x + 2)} + \frac{K \cdot 2}{2 (x + 2)}) + 1$

Step 7.2.9

Combine the numerators over the common denominator.

$y = \ln (\frac{5 (- 2 + x) + K \cdot 2}{2 (x + 2)}) + 1$

Step 7.2.10

Simplify the numerator.

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Step 7.2.10.1

Apply the distributive property.

$y = \ln (\frac{5 \cdot - 2 + 5 x + K \cdot 2}{2 (x + 2)}) + 1$

Step 7.2.10.2

Multiply $5$ by $- 2$ .

$y = \ln (\frac{- 10 + 5 x + K \cdot 2}{2 (x + 2)}) + 1$

Step 7.2.10.3

Move $2$ to the left of $K$ .

$y = \ln (\frac{- 10 + 5 x + 2 K}{2 (x + 2)}) + 1$