Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{{(2 + y)}^{2}}$ .

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

Step 1.2

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

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Step 1.2.1

Cancel the common factor.

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

Step 1.2.2

Rewrite the expression.

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Differentiate $2 + y$ .

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

Step 2.2.1.1.2

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

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

Step 2.2.1.1.3

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

$0 + \frac{d}{d y} [y]$

Step 2.2.1.1.4

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

$0 + 1$

Step 2.2.1.1.5

Add $0$ and $1$ .

$1$

Step 2.2.1.2

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

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

Step 2.2.2

Apply basic rules of exponents.

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Step 2.2.2.1

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

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

Step 2.2.2.2

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

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Step 2.2.2.2.1

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

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

Step 2.2.2.2.2

Multiply $2$ by $- 1$ .

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

Step 2.2.3

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

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

Step 2.2.4

Rewrite $- {u_{1}}^{- 1} + C_{1}$ as $- \frac{1}{u_{1}} + C_{1}$ .

$- \frac{1}{u_{1}} + C_{1} = \int \frac{1}{2 x - 1} d x$

Step 2.2.5

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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Step 2.3.1.1

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

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Step 2.3.1.1.1

Differentiate $2 x - 1$ .

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

Step 2.3.1.1.2

By the Sum Rule, the derivative of $2 x - 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$ .

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

Step 2.3.1.1.3

Evaluate $\frac{d}{d x} [2 x]$ .

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Step 2.3.1.1.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Step 2.3.1.1.3.2

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

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

Step 2.3.1.1.3.3

Multiply $2$ by $1$ .

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

Step 2.3.1.1.4

Differentiate using the Constant Rule.

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Step 2.3.1.1.4.1

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

$2 + 0$

Step 2.3.1.1.4.2

Add $2$ and $0$ .

$2$

Step 2.3.1.2

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

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

Step 2.3.2

Simplify.

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Step 2.3.2.1

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

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

Step 2.3.2.2

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

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

Simplify.

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

Step 2.3.6

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Simplify $\frac{1}{2} \ln (| 2 x - 1 |)$ by moving $\frac{1}{2}$ inside the logarithm.

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

Step 3.2

Find the LCD of the terms in the equation.

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Step 3.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$2 + y, 1, 1$

Step 3.2.2

Remove parentheses.

$2 + y, 1, 1$

Step 3.2.3

The LCM of one and any expression is the expression.

$2 + y$

Step 3.3

Multiply each term in $- \frac{1}{2 + y} = \ln ({| 2 x - 1 |}^{\frac{1}{2}}) + C$ by $2 + y$ to eliminate the fractions.

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Step 3.3.1

Multiply each term in $- \frac{1}{2 + y} = \ln ({| 2 x - 1 |}^{\frac{1}{2}}) + C$ by $2 + y$ .

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

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $2 + y$ .

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Step 3.3.2.1.1

Move the leading negative in $- \frac{1}{2 + y}$ into the numerator.

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

Step 3.3.2.1.2

Cancel the common factor.

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

Step 3.3.2.1.3

Rewrite the expression.

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

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Simplify each term.

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Step 3.3.3.1.1

Apply the distributive property.

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

Step 3.3.3.1.2

Multiply $\ln ({| 2 x - 1 |}^{\frac{1}{2}}) \cdot 2$ .

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Step 3.3.3.1.2.1

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

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

Step 3.3.3.1.2.2

Simplify $2 \cdot \ln ({| 2 x - 1 |}^{\frac{1}{2}})$ by moving $2$ inside the logarithm.

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

Step 3.3.3.1.3

Simplify each term.

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Step 3.3.3.1.3.1

Multiply the exponents in ${({| 2 x - 1 |}^{\frac{1}{2}})}^{2}$ .

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Step 3.3.3.1.3.1.1

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

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

Step 3.3.3.1.3.1.2

Cancel the common factor of $2$ .

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Step 3.3.3.1.3.1.2.1

Cancel the common factor.

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

Step 3.3.3.1.3.1.2.2

Rewrite the expression.

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

Step 3.3.3.1.3.2

Simplify.

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

Step 3.3.3.1.4

Apply the distributive property.

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

Step 3.3.3.1.5

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

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

Step 3.3.3.2

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

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

Step 3.4

Solve the equation.

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Step 3.4.1

Rewrite the equation as $\ln (| 2 x - 1 |) + y \ln ({| 2 x - 1 |}^{\frac{1}{2}}) + 2 C + C y = - 1$ .

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

Step 3.4.2

Move all the terms containing a logarithm to the left side of the equation.

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

Step 3.4.3

Add $C y$ to both sides of the equation.

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

Step 3.4.4

Subtract $\ln (| 2 x - 1 |)$ from both sides of the equation.

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

Step 3.4.5

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

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Step 3.4.5.1

Factor $y$ out of $C y$ .

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

Step 3.4.5.2

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

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

Step 3.4.6

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

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Step 3.4.6.1

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

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

Step 3.4.6.2

Simplify the left side.

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Step 3.4.6.2.1

Cancel the common factor of $\ln ({| 2 x - 1 |}^{\frac{1}{2}}) + C$ .

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Step 3.4.6.2.1.1

Cancel the common factor.

Step 3.4.6.2.1.2

Divide $y$ by $1$ .

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

Step 3.4.6.3

Simplify the right side.

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Step 3.4.6.3.1

Simplify each term.

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Step 3.4.6.3.1.1

Move the negative in front of the fraction.

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

Step 3.4.6.3.1.2

Move the negative in front of the fraction.

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

Step 3.4.6.3.1.3

Move the negative in front of the fraction.

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

Step 3.4.6.3.2

Simplify terms.

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Step 3.4.6.3.2.1

Combine the numerators over the common denominator.

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

Step 3.4.6.3.2.2

Combine the numerators over the common denominator.

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

Step 3.4.6.3.2.3

Factor $- 1$ out of $- 2 C$ .

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

Step 3.4.6.3.2.4

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

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

Step 3.4.6.3.2.5

Factor $- 1$ out of $- (2 C) - 1 (1)$ .

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

Step 3.4.6.3.2.6

Factor $- 1$ out of $- (2 C + 1) - \ln (| 2 x - 1 |)$ .

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

Step 3.4.6.3.2.7

Simplify the expression.

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Step 3.4.6.3.2.7.1

Rewrite $- (2 C + 1 + \ln (| 2 x - 1 |))$ as $- 1 (2 C + 1 + \ln (| 2 x - 1 |))$ .

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

Step 3.4.6.3.2.7.2

Move the negative in front of the fraction.

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

Step 4

Simplify the constant of integration.

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