Calculus Examples

$(y + 1) d x + (4 x - 1) d y = 0$

Step 1

Subtract $(y + 1) d x$ from both sides of the equation.

$(4 x - 1) d y = - (y + 1) d x$

Step 2

Multiply both sides by $\frac{1}{(4 x - 1) (y + 1)}$ .

$\frac{1}{(4 x - 1) (y + 1)} (4 x - 1) d y = \frac{1}{(4 x - 1) (y + 1)} (- (y + 1)) d x$

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $4 x - 1$ .

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Step 3.1.1

Cancel the common factor.

$\frac{1}{(4 x - 1) (y + 1)} (4 x - 1) d y = \frac{1}{(4 x - 1) (y + 1)} (- (y + 1)) d x$

Step 3.1.2

Rewrite the expression.

$\frac{1}{y + 1} d y = \frac{1}{(4 x - 1) (y + 1)} (- (y + 1)) d x$

Step 3.2

Rewrite using the commutative property of multiplication.

$\frac{1}{y + 1} d y = - \frac{1}{(4 x - 1) (y + 1)} (y + 1) d x$

Step 3.3

Cancel the common factor of $y + 1$ .

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Step 3.3.1

Move the leading negative in $- \frac{1}{(4 x - 1) (y + 1)}$ into the numerator.

$\frac{1}{y + 1} d y = \frac{- 1}{(4 x - 1) (y + 1)} (y + 1) d x$

Step 3.3.2

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

$\frac{1}{y + 1} d y = \frac{- 1}{(y + 1) (4 x - 1)} (y + 1) d x$

Step 3.3.3

Cancel the common factor.

$\frac{1}{y + 1} d y = \frac{- 1}{(y + 1) (4 x - 1)} (y + 1) d x$

Step 3.3.4

Rewrite the expression.

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

Step 3.4

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

$\int \frac{1}{y + 1} d y = \int - \frac{1}{4 x - 1} d x$

Step 4.2

Integrate the left side.

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Step 4.2.1

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

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Step 4.2.1.1

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

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Step 4.2.1.1.1

Differentiate $y + 1$ .

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

Step 4.2.1.1.2

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 4.2.1.1.3

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 4.2.1.1.4

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

$1 + 0$

Step 4.2.1.1.5

Add $1$ and $0$ .

$1$

Step 4.2.1.2

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

$\int \frac{1}{u_{1}} d u_{1} = \int - \frac{1}{4 x - 1} d x$

Step 4.2.2

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

$\ln (| u_{1} |) + C_{1} = \int - \frac{1}{4 x - 1} d x$

Step 4.2.3

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

$\ln (| y + 1 |) + C_{1} = \int - \frac{1}{4 x - 1} d x$

Step 4.3

Integrate the right side.

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Step 4.3.1

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

$\ln (| y + 1 |) + C_{1} = - \int \frac{1}{4 x - 1} d x$

Step 4.3.2

Let $u_{2} = 4 x - 1$ . Then $d u_{2} = 4 d x$ , so $\frac{1}{4} d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 4.3.2.1

Let $u_{2} = 4 x - 1$ . Find $\frac{d u_{2}}{d x}$ .

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Step 4.3.2.1.1

Differentiate $4 x - 1$ .

$\frac{d}{d x} [4 x - 1]$

Step 4.3.2.1.2

By the Sum Rule, the derivative of $4 x - 1$ with respect to $x$ is $\frac{d}{d x} [4 x] + \frac{d}{d x} [- 1]$ .

$\frac{d}{d x} [4 x] + \frac{d}{d x} [- 1]$

Step 4.3.2.1.3

Evaluate $\frac{d}{d x} [4 x]$ .

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Step 4.3.2.1.3.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$4 \frac{d}{d x} [x] + \frac{d}{d x} [- 1]$

Step 4.3.2.1.3.2

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

$4 \cdot 1 + \frac{d}{d x} [- 1]$

Step 4.3.2.1.3.3

Multiply $4$ by $1$ .

$4 + \frac{d}{d x} [- 1]$

Step 4.3.2.1.4

Differentiate using the Constant Rule.

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Step 4.3.2.1.4.1

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

$4 + 0$

Step 4.3.2.1.4.2

Add $4$ and $0$ .

$4$

Step 4.3.2.2

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

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

Step 4.3.3

Simplify.

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Step 4.3.3.1

Multiply $\frac{1}{u_{2}}$ by $\frac{1}{4}$ .

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

Step 4.3.3.2

Move $4$ to the left of $u_{2}$ .

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

Step 4.3.4

Since $\frac{1}{4}$ is constant with respect to $u_{2}$ , move $\frac{1}{4}$ out of the integral.

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

Step 4.3.5

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

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

Step 4.3.6

Simplify.

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

Step 4.3.7

Replace all occurrences of $u_{2}$ with $4 x - 1$ .

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

Step 4.4

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

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

Step 5

Solve for $y$ .

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Step 5.1

Simplify the right side.

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Step 5.1.1

Combine $\ln (| 4 x - 1 |)$ and $\frac{1}{4}$ .

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

Step 5.2

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

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

Step 5.3

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

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

Step 5.4

Simplify terms.

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Step 5.4.1

Combine $\ln (| y + 1 |)$ and $\frac{4}{4}$ .

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

Step 5.4.2

Combine the numerators over the common denominator.

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

Step 5.5

Move $4$ to the left of $\ln (| y + 1 |)$ .

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

Step 5.6

Simplify the left side.

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Step 5.6.1

Simplify $\frac{4 \ln (| y + 1 |) + \ln (| 4 x - 1 |)}{4}$ .

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Step 5.6.1.1

Simplify the numerator.

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Step 5.6.1.1.1

Simplify $4 \ln (| y + 1 |)$ by moving $4$ inside the logarithm.

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

Step 5.6.1.1.2

Remove the absolute value in ${| y + 1 |}^{4}$ because exponentiations with even powers are always positive.

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

Step 5.6.1.1.3

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

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

Step 5.6.1.2

Rewrite $\frac{\ln ({(y + 1)}^{4} | 4 x - 1 |)}{4}$ as $\frac{1}{4} \ln ({(y + 1)}^{4} | 4 x - 1 |)$ .

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

Step 5.6.1.3

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

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

Step 5.6.1.4

Apply the product rule to ${(y + 1)}^{4} | 4 x - 1 |$ .

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

Step 5.6.1.5

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

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Step 5.6.1.5.1

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

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

Step 5.6.1.5.2

Cancel the common factor of $4$ .

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Step 5.6.1.5.2.1

Cancel the common factor.

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

Step 5.6.1.5.2.2

Rewrite the expression.

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

Step 5.6.1.6

Simplify.

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

Step 5.6.1.7

Apply the distributive property.

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

Step 5.6.1.8

Multiply ${| 4 x - 1 |}^{\frac{1}{4}}$ by $1$ .

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

Step 5.7

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

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

Step 5.8

Rewrite $\ln (y {| 4 x - 1 |}^{\frac{1}{4}} + {| 4 x - 1 |}^{\frac{1}{4}}) = C$ 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 $b^{y} = x$ .

$e^{C} = y {| 4 x - 1 |}^{\frac{1}{4}} + {| 4 x - 1 |}^{\frac{1}{4}}$

Step 5.9

Solve for $y$ .

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Step 5.9.1

Rewrite the equation as $y {| 4 x - 1 |}^{\frac{1}{4}} + {| 4 x - 1 |}^{\frac{1}{4}} = e^{C}$ .

$y {| 4 x - 1 |}^{\frac{1}{4}} + {| 4 x - 1 |}^{\frac{1}{4}} = e^{C}$

Step 5.9.2

Subtract ${| 4 x - 1 |}^{\frac{1}{4}}$ from both sides of the equation.

$y {| 4 x - 1 |}^{\frac{1}{4}} = e^{C} - {| 4 x - 1 |}^{\frac{1}{4}}$

Step 5.9.3

Divide each term in $y {| 4 x - 1 |}^{\frac{1}{4}} = e^{C} - {| 4 x - 1 |}^{\frac{1}{4}}$ by ${| 4 x - 1 |}^{\frac{1}{4}}$ and simplify.

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Step 5.9.3.1

Divide each term in $y {| 4 x - 1 |}^{\frac{1}{4}} = e^{C} - {| 4 x - 1 |}^{\frac{1}{4}}$ by ${| 4 x - 1 |}^{\frac{1}{4}}$ .

$\frac{y {| 4 x - 1 |}^{\frac{1}{4}}}{{| 4 x - 1 |}^{\frac{1}{4}}} = \frac{e^{C}}{{| 4 x - 1 |}^{\frac{1}{4}}} + \frac{- {| 4 x - 1 |}^{\frac{1}{4}}}{{| 4 x - 1 |}^{\frac{1}{4}}}$

Step 5.9.3.2

Simplify the left side.

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Step 5.9.3.2.1

Cancel the common factor.

$\frac{y {| 4 x - 1 |}^{\frac{1}{4}}}{{| 4 x - 1 |}^{\frac{1}{4}}} = \frac{e^{C}}{{| 4 x - 1 |}^{\frac{1}{4}}} + \frac{- {| 4 x - 1 |}^{\frac{1}{4}}}{{| 4 x - 1 |}^{\frac{1}{4}}}$

Step 5.9.3.2.2

Divide $y$ by $1$ .

$y = \frac{e^{C}}{{| 4 x - 1 |}^{\frac{1}{4}}} + \frac{- {| 4 x - 1 |}^{\frac{1}{4}}}{{| 4 x - 1 |}^{\frac{1}{4}}}$

Step 5.9.3.3

Simplify the right side.

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Step 5.9.3.3.1

Combine the numerators over the common denominator.

$y = \frac{e^{C} - {| 4 x - 1 |}^{\frac{1}{4}}}{{| 4 x - 1 |}^{\frac{1}{4}}}$

Step 6

Simplify the constant of integration.

$y = \frac{C - {| 4 x - 1 |}^{\frac{1}{4}}}{{| 4 x - 1 |}^{\frac{1}{4}}}$