Calculus Examples

$\frac{d y}{d x} = \frac{\sqrt{y}}{2 x + 1}$

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{\sqrt{y}}$ .

$\frac{1}{\sqrt{y}} \frac{d y}{d x} = \frac{1}{\sqrt{y}} \cdot \frac{\sqrt{y}}{2 x + 1}$

Step 1.2

Cancel the common factor of $\sqrt{y}$ .

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Step 1.2.1

Cancel the common factor.

$\frac{1}{\sqrt{y}} \frac{d y}{d x} = \frac{1}{\sqrt{y}} \cdot \frac{\sqrt{y}}{2 x + 1}$

Step 1.2.2

Rewrite the expression.

$\frac{1}{\sqrt{y}} \frac{d y}{d x} = \frac{1}{2 x + 1}$

Step 1.3

Rewrite the equation.

$\frac{1}{\sqrt{y}} d y = \frac{1}{2 x + 1} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Apply basic rules of exponents.

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Step 2.2.1.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{y}$ as $y^{\frac{1}{2}}$ .

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

Step 2.2.1.2

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

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

Step 2.2.1.3

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

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Step 2.2.1.3.1

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

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

Step 2.2.1.3.2

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

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

Step 2.2.1.3.3

Move the negative in front of the fraction.

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

Step 2.2.2

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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Step 2.3.1.1

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

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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 $u$ and $d u$ .

$2 y^{\frac{1}{2}} + C_{1} = \int \frac{1}{u} \cdot \frac{1}{2} d u$

Step 2.3.2

Simplify.

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Step 2.3.2.1

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

$2 y^{\frac{1}{2}} + C_{1} = \int \frac{1}{u \cdot 2} d u$

Step 2.3.2.2

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

$2 y^{\frac{1}{2}} + C_{1} = \int \frac{1}{2 u} d u$

Step 2.3.3

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

$2 y^{\frac{1}{2}} + C_{1} = \frac{1}{2} \int \frac{1}{u} d u$

Step 2.3.4

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

$2 y^{\frac{1}{2}} + C_{1} = \frac{1}{2} (\ln (| u |) + C_{2})$

Step 2.3.5

Simplify.

$2 y^{\frac{1}{2}} + C_{1} = \frac{1}{2} \ln (| u |) + C_{2}$

Step 2.3.6

Replace all occurrences of $u$ with $2 x + 1$ .

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

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

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Step 3.1.1

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

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

Step 3.1.2

Simplify the left side.

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Step 3.1.2.1

Cancel the common factor.

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

Step 3.1.2.2

Divide $y^{\frac{1}{2}}$ by $1$ .

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

Step 3.1.3

Simplify the right side.

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Step 3.1.3.1

Simplify each term.

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Step 3.1.3.1.1

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

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

Step 3.1.3.1.2

Rewrite $\frac{\ln ({| 2 x + 1 |}^{\frac{1}{2}})}{2}$ as $\frac{1}{2} \ln ({| 2 x + 1 |}^{\frac{1}{2}})$ .

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

Step 3.1.3.1.3

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

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

Step 3.1.3.1.4

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

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Step 3.1.3.1.4.1

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

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

Step 3.1.3.1.4.2

Multiply $\frac{1}{2} \cdot \frac{1}{2}$ .

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Step 3.1.3.1.4.2.1

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

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

Step 3.1.3.1.4.2.2

Multiply $2$ by $2$ .

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

Step 3.1.3.2

To write $\ln ({| 2 x + 1 |}^{\frac{1}{4}})$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$y^{\frac{1}{2}} = \ln ({| 2 x + 1 |}^{\frac{1}{4}}) \cdot \frac{2}{2} + \frac{C}{2}$

Step 3.1.3.3

Simplify terms.

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Step 3.1.3.3.1

Combine $\ln ({| 2 x + 1 |}^{\frac{1}{4}})$ and $\frac{2}{2}$ .

$y^{\frac{1}{2}} = \frac{\ln ({| 2 x + 1 |}^{\frac{1}{4}}) \cdot 2}{2} + \frac{C}{2}$

Step 3.1.3.3.2

Combine the numerators over the common denominator.

$y^{\frac{1}{2}} = \frac{\ln ({| 2 x + 1 |}^{\frac{1}{4}}) \cdot 2 + C}{2}$

Step 3.1.3.4

Simplify the numerator.

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Step 3.1.3.4.1

Multiply $\ln ({| 2 x + 1 |}^{\frac{1}{4}}) \cdot 2$ .

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Step 3.1.3.4.1.1

Reorder $\ln ({| 2 x + 1 |}^{\frac{1}{4}})$ and $2$ .

$y^{\frac{1}{2}} = \frac{2 \cdot \ln ({| 2 x + 1 |}^{\frac{1}{4}}) + C}{2}$

Step 3.1.3.4.1.2

Simplify $2 \cdot \ln ({| 2 x + 1 |}^{\frac{1}{4}})$ by moving $2$ inside the logarithm.

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

Step 3.1.3.4.2

Multiply the exponents in ${({| 2 x + 1 |}^{\frac{1}{4}})}^{2}$ .

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Step 3.1.3.4.2.1

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

$y^{\frac{1}{2}} = \frac{\ln ({| 2 x + 1 |}^{\frac{1}{4} \cdot 2}) + C}{2}$

Step 3.1.3.4.2.2

Cancel the common factor of $2$ .

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Step 3.1.3.4.2.2.1

Factor $2$ out of $4$ .

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

Step 3.1.3.4.2.2.2

Cancel the common factor.

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

Step 3.1.3.4.2.2.3

Rewrite the expression.

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

Step 3.2

Raise each side of the equation to the power of $2$ to eliminate the fractional exponent on the left side.

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

Step 3.3

Simplify the exponent.

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Step 3.3.1

Simplify the left side.

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Step 3.3.1.1

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

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Step 3.3.1.1.1

Multiply the exponents in ${(y^{\frac{1}{2}})}^{2}$ .

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Step 3.3.1.1.1.1

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

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

Step 3.3.1.1.1.2

Cancel the common factor of $2$ .

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Step 3.3.1.1.1.2.1

Cancel the common factor.

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

Step 3.3.1.1.1.2.2

Rewrite the expression.

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

Step 3.3.1.1.2

Simplify.

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

Step 3.3.2

Simplify the right side.

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Step 3.3.2.1

Simplify ${(\frac{\ln ({| 2 x + 1 |}^{\frac{1}{2}}) + C}{2})}^{2}$ .

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Step 3.3.2.1.1

Split the fraction $\frac{\ln ({| 2 x + 1 |}^{\frac{1}{2}}) + C}{2}$ into two fractions.

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

Step 3.3.2.1.2

Simplify each term.

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Step 3.3.2.1.2.1

Rewrite $\frac{\ln ({| 2 x + 1 |}^{\frac{1}{2}})}{2}$ as $\frac{1}{2} \ln ({| 2 x + 1 |}^{\frac{1}{2}})$ .

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

Step 3.3.2.1.2.2

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

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

Step 3.3.2.1.2.3

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

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Step 3.3.2.1.2.3.1

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

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

Step 3.3.2.1.2.3.2

Multiply $\frac{1}{2} \cdot \frac{1}{2}$ .

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Step 3.3.2.1.2.3.2.1

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

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

Step 3.3.2.1.2.3.2.2

Multiply $2$ by $2$ .

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

Step 4

Simplify the constant of integration.

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