Calculus Examples

$(1 + x^{2}) d y - x y d x = 0$

Step 1

Add $x y d x$ to both sides of the equation.

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

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

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Step 3.1.1

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

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

Step 3.1.2

Cancel the common factor.

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

Step 3.1.3

Rewrite the expression.

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

Step 3.2

Cancel the common factor of $y$ .

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Step 3.2.1

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

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

Step 3.2.2

Factor $y$ out of $x y$ .

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

Step 3.2.3

Cancel the common factor.

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

Step 3.2.4

Rewrite the expression.

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

Step 3.3

Combine $\frac{1}{1 + x^{2}}$ and $x$ .

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

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

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

Step 4.3

Integrate the right side.

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Step 4.3.1

Let $u = 1 + x^{2}$ . 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 4.3.1.1

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

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Step 4.3.1.1.1

Differentiate $1 + x^{2}$ .

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

Step 4.3.1.1.2

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

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

Step 4.3.1.1.3

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

$0 + \frac{d}{d x} [x^{2}]$

Step 4.3.1.1.4

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

$0 + 2 x$

Step 4.3.1.1.5

Add $0$ and $2 x$ .

$2 x$

Step 4.3.1.2

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

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

Step 4.3.2

Simplify.

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Step 4.3.2.1

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

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

Step 4.3.2.2

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

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

Step 4.3.3

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

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

Step 4.3.4

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

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

Step 4.3.5

Simplify.

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

Step 4.3.6

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

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

Step 4.4

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

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

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

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

Step 5.2

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

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

Step 5.3

To write $\ln (| y |)$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 5.4

Simplify terms.

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Step 5.4.1

Combine $\ln (| y |)$ and $\frac{2}{2}$ .

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

Step 5.4.2

Combine the numerators over the common denominator.

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

Step 5.5

Move $2$ to the left of $\ln (| y |)$ .

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

Step 5.6

Simplify the left side.

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Step 5.6.1

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

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Step 5.6.1.1

Simplify the numerator.

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Step 5.6.1.1.1

Simplify $2 \ln (| y |)$ by moving $2$ inside the logarithm.

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

Step 5.6.1.1.2

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

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

Step 5.6.1.1.3

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

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

Step 5.6.1.2

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

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

Step 5.6.1.3

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

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

Step 5.6.1.4

Apply the product rule to $\frac{y^{2}}{| 1 + x^{2} |}$ .

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

Step 5.6.1.5

Simplify the numerator.

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Step 5.6.1.5.1

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

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Step 5.6.1.5.1.1

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

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

Step 5.6.1.5.1.2

Cancel the common factor of $2$ .

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Step 5.6.1.5.1.2.1

Cancel the common factor.

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

Step 5.6.1.5.1.2.2

Rewrite the expression.

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

Step 5.6.1.5.2

Simplify.

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

Step 5.7

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

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

Step 5.8

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

$e^{C} = \frac{y}{{| 1 + x^{2} |}^{\frac{1}{2}}}$

Step 5.9

Solve for $y$ .

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Step 5.9.1

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

$\frac{y}{{| 1 + x^{2} |}^{\frac{1}{2}}} = e^{C}$

Step 5.9.2

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

$\frac{y}{{| 1 + x^{2} |}^{\frac{1}{2}}} {| 1 + x^{2} |}^{\frac{1}{2}} = e^{C} {| 1 + x^{2} |}^{\frac{1}{2}}$

Step 5.9.3

Simplify the left side.

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Step 5.9.3.1

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

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Step 5.9.3.1.1

Cancel the common factor.

$\frac{y}{{| 1 + x^{2} |}^{\frac{1}{2}}} {| 1 + x^{2} |}^{\frac{1}{2}} = e^{C} {| 1 + x^{2} |}^{\frac{1}{2}}$

Step 5.9.3.1.2

Rewrite the expression.

$y = e^{C} {| 1 + x^{2} |}^{\frac{1}{2}}$

Step 6

Simplify the constant of integration.

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