Calculus Examples

$4 t x \frac{d x}{d t} = x^{2} + 1$

Step 1

Separate the variables.

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Step 1.1

Divide each term in $4 t x \frac{d x}{d t} = x^{2} + 1$ by $4 t x$ and simplify.

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Step 1.1.1

Divide each term in $4 t x \frac{d x}{d t} = x^{2} + 1$ by $4 t x$ .

$\frac{4 t x \frac{d x}{d t}}{4 t x} = \frac{x^{2}}{4 t x} + \frac{1}{4 t x}$

Step 1.1.2

Simplify the left side.

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Step 1.1.2.1

Cancel the common factor of $4$ .

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Step 1.1.2.1.1

Cancel the common factor.

$\frac{4 t x \frac{d x}{d t}}{4 t x} = \frac{x^{2}}{4 t x} + \frac{1}{4 t x}$

Step 1.1.2.1.2

Rewrite the expression.

$\frac{t x \frac{d x}{d t}}{t x} = \frac{x^{2}}{4 t x} + \frac{1}{4 t x}$

Step 1.1.2.2

Cancel the common factor of $t$ .

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Step 1.1.2.2.1

Cancel the common factor.

$\frac{t x \frac{d x}{d t}}{t x} = \frac{x^{2}}{4 t x} + \frac{1}{4 t x}$

Step 1.1.2.2.2

Rewrite the expression.

$\frac{x \frac{d x}{d t}}{x} = \frac{x^{2}}{4 t x} + \frac{1}{4 t x}$

Step 1.1.2.3

Cancel the common factor of $x$ .

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Step 1.1.2.3.1

Cancel the common factor.

$\frac{x \frac{d x}{d t}}{x} = \frac{x^{2}}{4 t x} + \frac{1}{4 t x}$

Step 1.1.2.3.2

Divide $\frac{d x}{d t}$ by $1$ .

$\frac{d x}{d t} = \frac{x^{2}}{4 t x} + \frac{1}{4 t x}$

Step 1.1.3

Simplify the right side.

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Step 1.1.3.1

Cancel the common factor of $x^{2}$ and $x$ .

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Step 1.1.3.1.1

Factor $x$ out of $x^{2}$ .

$\frac{d x}{d t} = \frac{x \cdot x}{4 t x} + \frac{1}{4 t x}$

Step 1.1.3.1.2

Cancel the common factors.

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Step 1.1.3.1.2.1

Factor $x$ out of $4 t x$ .

$\frac{d x}{d t} = \frac{x \cdot x}{x (4 t)} + \frac{1}{4 t x}$

Step 1.1.3.1.2.2

Cancel the common factor.

$\frac{d x}{d t} = \frac{x \cdot x}{x (4 t)} + \frac{1}{4 t x}$

Step 1.1.3.1.2.3

Rewrite the expression.

$\frac{d x}{d t} = \frac{x}{4 t} + \frac{1}{4 t x}$

Step 1.2

Factor.

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Step 1.2.1

To write $\frac{x}{4 t}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$\frac{d x}{d t} = \frac{x}{4 t} \cdot \frac{x}{x} + \frac{1}{4 t x}$

Step 1.2.2

Multiply $\frac{x}{4 t}$ by $\frac{x}{x}$ .

$\frac{d x}{d t} = \frac{x \cdot x}{4 t x} + \frac{1}{4 t x}$

Step 1.2.3

Combine the numerators over the common denominator.

$\frac{d x}{d t} = \frac{x \cdot x + 1}{4 t x}$

Step 1.2.4

Multiply $x$ by $x$ .

$\frac{d x}{d t} = \frac{x^{2} + 1}{4 t x}$

Step 1.3

Regroup factors.

$\frac{d x}{d t} = \frac{x^{2} + 1}{4 x} \cdot \frac{1}{t}$

Step 1.4

Multiply both sides by $\frac{4 x}{x^{2} + 1}$ .

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

Step 1.5

Simplify.

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Step 1.5.1

Multiply $\frac{x^{2} + 1}{4 x}$ by $\frac{1}{t}$ .

$\frac{4 x}{x^{2} + 1} \frac{d x}{d t} = \frac{4 x}{x^{2} + 1} \cdot \frac{x^{2} + 1}{4 x t}$

Step 1.5.2

Cancel the common factor of $4 x$ .

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Step 1.5.2.1

Factor $4 x$ out of $4 x t$ .

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

Step 1.5.2.2

Cancel the common factor.

$\frac{4 x}{x^{2} + 1} \frac{d x}{d t} = \frac{4 x}{x^{2} + 1} \cdot \frac{x^{2} + 1}{4 x t}$

Step 1.5.2.3

Rewrite the expression.

$\frac{4 x}{x^{2} + 1} \frac{d x}{d t} = \frac{1}{x^{2} + 1} \cdot \frac{x^{2} + 1}{t}$

Step 1.5.3

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

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Step 1.5.3.1

Cancel the common factor.

$\frac{4 x}{x^{2} + 1} \frac{d x}{d t} = \frac{1}{x^{2} + 1} \cdot \frac{x^{2} + 1}{t}$

Step 1.5.3.2

Rewrite the expression.

$\frac{4 x}{x^{2} + 1} \frac{d x}{d t} = \frac{1}{t}$

Step 1.6

Rewrite the equation.

$\frac{4 x}{x^{2} + 1} d x = \frac{1}{t} d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{4 x}{x^{2} + 1} d x = \int \frac{1}{t} d t$

Step 2.2

Integrate the left side.

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Step 2.2.1

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

$4 \int \frac{x}{x^{2} + 1} d x = \int \frac{1}{t} d t$

Step 2.2.2

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

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Step 2.2.2.1

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

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Step 2.2.2.1.1

Differentiate $x^{2} + 1$ .

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

Step 2.2.2.1.2

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

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

Step 2.2.2.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

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

Step 2.2.2.1.4

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

$2 x + 0$

Step 2.2.2.1.5

Add $2 x$ and $0$ .

$2 x$

Step 2.2.2.2

Rewrite the problem using $u$ and $d u$ .

$4 \int \frac{1}{u} \cdot \frac{1}{2} d u = \int \frac{1}{t} d t$

Step 2.2.3

Simplify.

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Step 2.2.3.1

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

$4 \int \frac{1}{u \cdot 2} d u = \int \frac{1}{t} d t$

Step 2.2.3.2

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

$4 \int \frac{1}{2 u} d u = \int \frac{1}{t} d t$

Step 2.2.4

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

$4 (\frac{1}{2} \int \frac{1}{u} d u) = \int \frac{1}{t} d t$

Step 2.2.5

Simplify.

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Step 2.2.5.1

Combine $\frac{1}{2}$ and $4$ .

$\frac{4}{2} \int \frac{1}{u} d u = \int \frac{1}{t} d t$

Step 2.2.5.2

Cancel the common factor of $4$ and $2$ .

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Step 2.2.5.2.1

Factor $2$ out of $4$ .

$\frac{2 \cdot 2}{2} \int \frac{1}{u} d u = \int \frac{1}{t} d t$

Step 2.2.5.2.2

Cancel the common factors.

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Step 2.2.5.2.2.1

Factor $2$ out of $2$ .

$\frac{2 \cdot 2}{2 (1)} \int \frac{1}{u} d u = \int \frac{1}{t} d t$

Step 2.2.5.2.2.2

Cancel the common factor.

$\frac{2 \cdot 2}{2 \cdot 1} \int \frac{1}{u} d u = \int \frac{1}{t} d t$

Step 2.2.5.2.2.3

Rewrite the expression.

$\frac{2}{1} \int \frac{1}{u} d u = \int \frac{1}{t} d t$

Step 2.2.5.2.2.4

Divide $2$ by $1$ .

$2 \int \frac{1}{u} d u = \int \frac{1}{t} d t$

Step 2.2.6

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

$2 (\ln (| u |) + C_{1}) = \int \frac{1}{t} d t$

Step 2.2.7

Simplify.

$2 \ln (| u |) + C_{1} = \int \frac{1}{t} d t$

Step 2.2.8

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 3

Solve for $x$ .

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Step 3.1

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

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

Step 3.2

Simplify the left side.

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Step 3.2.1

Simplify $2 \ln (| x^{2} + 1 |) - \ln (| t |)$ .

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Step 3.2.1.1

Simplify each term.

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Step 3.2.1.1.1

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

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

Step 3.2.1.1.2

Remove the absolute value in ${| x^{2} + 1 |}^{2}$ because exponentiations with even powers are always positive.

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

Step 3.2.1.2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

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

Step 3.3

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

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

Step 3.4

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

$e^{C} = \frac{{(x^{2} + 1)}^{2}}{| t |}$

Step 3.5

Solve for $x$ .

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Step 3.5.1

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

$\frac{{(x^{2} + 1)}^{2}}{| t |} = e^{C}$

Step 3.5.2

Multiply both sides by $| t |$ .

$\frac{{(x^{2} + 1)}^{2}}{| t |} | t | = e^{C} | t |$

Step 3.5.3

Simplify the left side.

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Step 3.5.3.1

Cancel the common factor of $| t |$ .

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Step 3.5.3.1.1

Cancel the common factor.

$\frac{{(x^{2} + 1)}^{2}}{| t |} | t | = e^{C} | t |$

Step 3.5.3.1.2

Rewrite the expression.

${(x^{2} + 1)}^{2} = e^{C} | t |$

Step 3.5.4

Solve for $x$ .

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Step 3.5.4.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x^{2} + 1 = \pm \sqrt{e^{C} | t |}$

Step 3.5.4.2

The complete solution is the result of both the positive and negative portions of the solution.

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Step 3.5.4.2.1

First, use the positive value of the $\pm$ to find the first solution.

$x^{2} + 1 = \sqrt{e^{C} | t |}$

Step 3.5.4.2.2

Subtract $1$ from both sides of the equation.

$x^{2} = \sqrt{e^{C} | t |} - 1$

Step 3.5.4.2.3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{\sqrt{e^{C} | t |} - 1}$

Step 3.5.4.2.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 3.5.4.2.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{\sqrt{e^{C} | t |} - 1}$

Step 3.5.4.2.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{\sqrt{e^{C} | t |} - 1}$

Step 3.5.4.2.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{\sqrt{e^{C} | t |} - 1}$

$x = - \sqrt{\sqrt{e^{C} | t |} - 1}$

$x = \sqrt{\sqrt{e^{C} | t |} - 1}$

$x = - \sqrt{\sqrt{e^{C} | t |} - 1}$

Step 3.5.4.2.5

Next, use the negative value of the $\pm$ to find the second solution.

$x^{2} + 1 = - \sqrt{e^{C} | t |}$

Step 3.5.4.2.6

Subtract $1$ from both sides of the equation.

$x^{2} = - \sqrt{e^{C} | t |} - 1$

Step 3.5.4.2.7

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- \sqrt{e^{C} | t |} - 1}$

Step 3.5.4.2.8

The complete solution is the result of both the positive and negative portions of the solution.

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Step 3.5.4.2.8.1

First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{- \sqrt{e^{C} | t |} - 1}$

Step 3.5.4.2.8.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{- \sqrt{e^{C} | t |} - 1}$

Step 3.5.4.2.8.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{- \sqrt{e^{C} | t |} - 1}$

$x = - \sqrt{- \sqrt{e^{C} | t |} - 1}$

$x = \sqrt{- \sqrt{e^{C} | t |} - 1}$

$x = - \sqrt{- \sqrt{e^{C} | t |} - 1}$

Step 3.5.4.2.9

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{\sqrt{e^{C} | t |} - 1}$

$x = - \sqrt{\sqrt{e^{C} | t |} - 1}$

$x = \sqrt{- \sqrt{e^{C} | t |} - 1}$

$x = - \sqrt{- \sqrt{e^{C} | t |} - 1}$

$x = \sqrt{\sqrt{e^{C} | t |} - 1}$

$x = - \sqrt{\sqrt{e^{C} | t |} - 1}$

$x = \sqrt{- \sqrt{e^{C} | t |} - 1}$

$x = - \sqrt{- \sqrt{e^{C} | t |} - 1}$

$x = \sqrt{\sqrt{e^{C} | t |} - 1}$

$x = - \sqrt{\sqrt{e^{C} | t |} - 1}$

$x = \sqrt{- \sqrt{e^{C} | t |} - 1}$

$x = - \sqrt{- \sqrt{e^{C} | t |} - 1}$

$x = \sqrt{\sqrt{e^{C} | t |} - 1}$

$x = - \sqrt{\sqrt{e^{C} | t |} - 1}$

$x = \sqrt{- \sqrt{e^{C} | t |} - 1}$

$x = - \sqrt{- \sqrt{e^{C} | t |} - 1}$

$x = \sqrt{\sqrt{e^{C} | t |} - 1}$

$x = - \sqrt{\sqrt{e^{C} | t |} - 1}$

$x = \sqrt{- \sqrt{e^{C} | t |} - 1}$

$x = - \sqrt{- \sqrt{e^{C} | t |} - 1}$

Step 4

Simplify the constant of integration.

$x = \sqrt{\sqrt{C | t |} - 1}$

$x = - \sqrt{\sqrt{C | t |} - 1}$

$x = \sqrt{- \sqrt{C | t |} - 1}$

$x = - \sqrt{- \sqrt{C | t |} - 1}$