Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Solve for $\frac{d y}{d x}$ .

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Step 1.1.1

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

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

Step 1.1.2

Multiply both sides by $2 + y$ .

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

Step 1.1.3

Simplify.

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Step 1.1.3.1

Simplify the left side.

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Step 1.1.3.1.1

Cancel the common factor of $2 + y$ .

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Step 1.1.3.1.1.1

Cancel the common factor.

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

Step 1.1.3.1.1.2

Rewrite the expression.

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

Step 1.1.3.2

Simplify the right side.

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Step 1.1.3.2.1

Simplify $- 2 x (2 + y)$ .

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Step 1.1.3.2.1.1

Apply the distributive property.

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

Step 1.1.3.2.1.2

Simplify the expression.

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Step 1.1.3.2.1.2.1

Multiply $2$ by $- 2$ .

$\sqrt{1 + x^{2}} \frac{d y}{d x} = - 4 x - 2 x y$

Step 1.1.3.2.1.2.2

Reorder $- 4 x$ and $- 2 x y$ .

$\sqrt{1 + x^{2}} \frac{d y}{d x} = - 2 x y - 4 x$

Step 1.1.4

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

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Step 1.1.4.1

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

$\frac{\sqrt{1 + x^{2}} \frac{d y}{d x}}{\sqrt{1 + x^{2}}} = \frac{- 2 x y}{\sqrt{1 + x^{2}}} + \frac{- 4 x}{\sqrt{1 + x^{2}}}$

Step 1.1.4.2

Simplify the left side.

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Step 1.1.4.2.1

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

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Step 1.1.4.2.1.1

Cancel the common factor.

$\frac{\sqrt{1 + x^{2}} \frac{d y}{d x}}{\sqrt{1 + x^{2}}} = \frac{- 2 x y}{\sqrt{1 + x^{2}}} + \frac{- 4 x}{\sqrt{1 + x^{2}}}$

Step 1.1.4.2.1.2

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

$\frac{d y}{d x} = \frac{- 2 x y}{\sqrt{1 + x^{2}}} + \frac{- 4 x}{\sqrt{1 + x^{2}}}$

Step 1.1.4.3

Simplify the right side.

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Step 1.1.4.3.1

Simplify terms.

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Step 1.1.4.3.1.1

Simplify each term.

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Step 1.1.4.3.1.1.1

Move the negative in front of the fraction.

$\frac{d y}{d x} = - \frac{2 x y}{\sqrt{1 + x^{2}}} + \frac{- 4 x}{\sqrt{1 + x^{2}}}$

Step 1.1.4.3.1.1.2

Multiply $\frac{2 x y}{\sqrt{1 + x^{2}}}$ by $\frac{\sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}}$ .

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

Step 1.1.4.3.1.1.3

Combine and simplify the denominator.

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Step 1.1.4.3.1.1.3.1

Multiply $\frac{2 x y}{\sqrt{1 + x^{2}}}$ by $\frac{\sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}}$ .

$\frac{d y}{d x} = - \frac{2 x y \sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}} \sqrt{1 + x^{2}}} + \frac{- 4 x}{\sqrt{1 + x^{2}}}$

Step 1.1.4.3.1.1.3.2

Raise $\sqrt{1 + x^{2}}$ to the power of $1$ .

$\frac{d y}{d x} = - \frac{2 x y \sqrt{1 + x^{2}}}{{\sqrt{1 + x^{2}}}^{1} \sqrt{1 + x^{2}}} + \frac{- 4 x}{\sqrt{1 + x^{2}}}$

Step 1.1.4.3.1.1.3.3

Raise $\sqrt{1 + x^{2}}$ to the power of $1$ .

$\frac{d y}{d x} = - \frac{2 x y \sqrt{1 + x^{2}}}{{\sqrt{1 + x^{2}}}^{1} {\sqrt{1 + x^{2}}}^{1}} + \frac{- 4 x}{\sqrt{1 + x^{2}}}$

Step 1.1.4.3.1.1.3.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{d y}{d x} = - \frac{2 x y \sqrt{1 + x^{2}}}{{\sqrt{1 + x^{2}}}^{1 + 1}} + \frac{- 4 x}{\sqrt{1 + x^{2}}}$

Step 1.1.4.3.1.1.3.5

Add $1$ and $1$ .

$\frac{d y}{d x} = - \frac{2 x y \sqrt{1 + x^{2}}}{{\sqrt{1 + x^{2}}}^{2}} + \frac{- 4 x}{\sqrt{1 + x^{2}}}$

Step 1.1.4.3.1.1.3.6

Rewrite ${\sqrt{1 + x^{2}}}^{2}$ as $1 + x^{2}$ .

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Step 1.1.4.3.1.1.3.6.1

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

$\frac{d y}{d x} = - \frac{2 x y \sqrt{1 + x^{2}}}{{({(1 + x^{2})}^{\frac{1}{2}})}^{2}} + \frac{- 4 x}{\sqrt{1 + x^{2}}}$

Step 1.1.4.3.1.1.3.6.2

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

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

Step 1.1.4.3.1.1.3.6.3

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

$\frac{d y}{d x} = - \frac{2 x y \sqrt{1 + x^{2}}}{{(1 + x^{2})}^{\frac{2}{2}}} + \frac{- 4 x}{\sqrt{1 + x^{2}}}$

Step 1.1.4.3.1.1.3.6.4

Cancel the common factor of $2$ .

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Step 1.1.4.3.1.1.3.6.4.1

Cancel the common factor.

$\frac{d y}{d x} = - \frac{2 x y \sqrt{1 + x^{2}}}{{(1 + x^{2})}^{\frac{2}{2}}} + \frac{- 4 x}{\sqrt{1 + x^{2}}}$

Step 1.1.4.3.1.1.3.6.4.2

Rewrite the expression.

$\frac{d y}{d x} = - \frac{2 x y \sqrt{1 + x^{2}}}{{(1 + x^{2})}^{1}} + \frac{- 4 x}{\sqrt{1 + x^{2}}}$

Step 1.1.4.3.1.1.3.6.5

Simplify.

$\frac{d y}{d x} = - \frac{2 x y \sqrt{1 + x^{2}}}{1 + x^{2}} + \frac{- 4 x}{\sqrt{1 + x^{2}}}$

Step 1.1.4.3.1.1.4

Move the negative in front of the fraction.

$\frac{d y}{d x} = - \frac{2 x y \sqrt{1 + x^{2}}}{1 + x^{2}} - \frac{4 x}{\sqrt{1 + x^{2}}}$

Step 1.1.4.3.1.1.5

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

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

Step 1.1.4.3.1.1.6

Combine and simplify the denominator.

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Step 1.1.4.3.1.1.6.1

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

$\frac{d y}{d x} = - \frac{2 x y \sqrt{1 + x^{2}}}{1 + x^{2}} - \frac{4 x \sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}} \sqrt{1 + x^{2}}}$

Step 1.1.4.3.1.1.6.2

Raise $\sqrt{1 + x^{2}}$ to the power of $1$ .

$\frac{d y}{d x} = - \frac{2 x y \sqrt{1 + x^{2}}}{1 + x^{2}} - \frac{4 x \sqrt{1 + x^{2}}}{{\sqrt{1 + x^{2}}}^{1} \sqrt{1 + x^{2}}}$

Step 1.1.4.3.1.1.6.3

Raise $\sqrt{1 + x^{2}}$ to the power of $1$ .

$\frac{d y}{d x} = - \frac{2 x y \sqrt{1 + x^{2}}}{1 + x^{2}} - \frac{4 x \sqrt{1 + x^{2}}}{{\sqrt{1 + x^{2}}}^{1} {\sqrt{1 + x^{2}}}^{1}}$

Step 1.1.4.3.1.1.6.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{d y}{d x} = - \frac{2 x y \sqrt{1 + x^{2}}}{1 + x^{2}} - \frac{4 x \sqrt{1 + x^{2}}}{{\sqrt{1 + x^{2}}}^{1 + 1}}$

Step 1.1.4.3.1.1.6.5

Add $1$ and $1$ .

$\frac{d y}{d x} = - \frac{2 x y \sqrt{1 + x^{2}}}{1 + x^{2}} - \frac{4 x \sqrt{1 + x^{2}}}{{\sqrt{1 + x^{2}}}^{2}}$

Step 1.1.4.3.1.1.6.6

Rewrite ${\sqrt{1 + x^{2}}}^{2}$ as $1 + x^{2}$ .

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Step 1.1.4.3.1.1.6.6.1

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

$\frac{d y}{d x} = - \frac{2 x y \sqrt{1 + x^{2}}}{1 + x^{2}} - \frac{4 x \sqrt{1 + x^{2}}}{{({(1 + x^{2})}^{\frac{1}{2}})}^{2}}$

Step 1.1.4.3.1.1.6.6.2

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

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

Step 1.1.4.3.1.1.6.6.3

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

$\frac{d y}{d x} = - \frac{2 x y \sqrt{1 + x^{2}}}{1 + x^{2}} - \frac{4 x \sqrt{1 + x^{2}}}{{(1 + x^{2})}^{\frac{2}{2}}}$

Step 1.1.4.3.1.1.6.6.4

Cancel the common factor of $2$ .

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Step 1.1.4.3.1.1.6.6.4.1

Cancel the common factor.

$\frac{d y}{d x} = - \frac{2 x y \sqrt{1 + x^{2}}}{1 + x^{2}} - \frac{4 x \sqrt{1 + x^{2}}}{{(1 + x^{2})}^{\frac{2}{2}}}$

Step 1.1.4.3.1.1.6.6.4.2

Rewrite the expression.

$\frac{d y}{d x} = - \frac{2 x y \sqrt{1 + x^{2}}}{1 + x^{2}} - \frac{4 x \sqrt{1 + x^{2}}}{{(1 + x^{2})}^{1}}$

Step 1.1.4.3.1.1.6.6.5

Simplify.

$\frac{d y}{d x} = - \frac{2 x y \sqrt{1 + x^{2}}}{1 + x^{2}} - \frac{4 x \sqrt{1 + x^{2}}}{1 + x^{2}}$

Step 1.1.4.3.1.2

Combine the numerators over the common denominator.

$\frac{d y}{d x} = \frac{- 2 x y \sqrt{1 + x^{2}} - 4 x \sqrt{1 + x^{2}}}{1 + x^{2}}$

Step 1.1.4.3.2

Simplify the numerator.

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Step 1.1.4.3.2.1

Factor $2 x \sqrt{1 + x^{2}}$ out of $- 2 x y \sqrt{1 + x^{2}} - 4 x \sqrt{1 + x^{2}}$ .

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Step 1.1.4.3.2.1.1

Factor $2 x \sqrt{1 + x^{2}}$ out of $- 2 x y \sqrt{1 + x^{2}}$ .

$\frac{d y}{d x} = \frac{2 x \sqrt{1 + x^{2}} (- 1 y) - 4 x \sqrt{1 + x^{2}}}{1 + x^{2}}$

Step 1.1.4.3.2.1.2

Factor $2 x \sqrt{1 + x^{2}}$ out of $- 4 x \sqrt{1 + x^{2}}$ .

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

Step 1.1.4.3.2.1.3

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

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

Step 1.1.4.3.2.2

Rewrite $- 1 y$ as $- y$ .

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

Step 1.1.4.3.3

Simplify with factoring out.

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Step 1.1.4.3.3.1

Factor $- 1$ out of $- y$ .

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

Step 1.1.4.3.3.2

Rewrite $- 2$ as $- 1 (2)$ .

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

Step 1.1.4.3.3.3

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

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

Step 1.1.4.3.3.4

Simplify the expression.

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Step 1.1.4.3.3.4.1

Rewrite $- (y + 2)$ as $- 1 (y + 2)$ .

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

Step 1.1.4.3.3.4.2

Move the negative in front of the fraction.

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

Step 1.2

Regroup factors.

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

Step 1.3

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

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

Step 1.4

Simplify.

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Step 1.4.1

Rewrite using the commutative property of multiplication.

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

Step 1.4.2

Cancel the common factor of $y + 2$ .

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Step 1.4.2.1

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

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

Step 1.4.2.2

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

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

Step 1.4.2.3

Cancel the common factor.

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

Step 1.4.2.4

Rewrite the expression.

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Differentiate $y + 2$ .

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

Step 2.2.1.1.2

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

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

Step 2.2.1.1.3

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

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

Step 2.2.1.1.4

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

$1 + 0$

Step 2.2.1.1.5

Add $1$ and $0$ .

$1$

Step 2.2.1.2

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

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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

Step 2.3.2

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

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

Step 2.3.3

Multiply $2$ by $- 1$ .

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

Step 2.3.4

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

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Step 2.3.4.1

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

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Step 2.3.4.1.1

Differentiate $1 + x^{2}$ .

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

Step 2.3.4.1.2

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 2.3.4.1.3

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 2.3.4.1.4

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 2.3.4.1.5

Add $0$ and $2 x$ .

$2 x$

Step 2.3.4.2

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

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

Step 2.3.5

Simplify.

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Step 2.3.5.1

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

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

Step 2.3.5.2

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

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

Step 2.3.6

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

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

Step 2.3.7

Simplify the expression.

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Step 2.3.7.1

Simplify.

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Step 2.3.7.1.1

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

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

Step 2.3.7.1.2

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

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Step 2.3.7.1.2.1

Factor $2$ out of $- 2$ .

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

Step 2.3.7.1.2.2

Cancel the common factors.

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Step 2.3.7.1.2.2.1

Factor $2$ out of $2$ .

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

Step 2.3.7.1.2.2.2

Cancel the common factor.

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

Step 2.3.7.1.2.2.3

Rewrite the expression.

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

Step 2.3.7.1.2.2.4

Divide $- 1$ by $1$ .

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

Step 2.3.7.2

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

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

Step 2.3.7.3

Simplify.

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Step 2.3.7.3.1

Move ${u_{2}}^{\frac{1}{2}}$ to the denominator using the negative exponent rule $b^{n} = \frac{1}{b^{- n}}$ .

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

Step 2.3.7.3.2

Multiply $u_{2}$ by ${u_{2}}^{- \frac{1}{2}}$ by adding the exponents.

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Step 2.3.7.3.2.1

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

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Step 2.3.7.3.2.1.1

Raise $u_{2}$ to the power of $1$ .

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

Step 2.3.7.3.2.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 2.3.7.3.2.2

Write $1$ as a fraction with a common denominator.

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

Step 2.3.7.3.2.3

Combine the numerators over the common denominator.

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

Step 2.3.7.3.2.4

Subtract $1$ from $2$ .

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

Step 2.3.7.4

Apply basic rules of exponents.

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Step 2.3.7.4.1

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

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

Step 2.3.7.4.2

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

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Step 2.3.7.4.2.1

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

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

Step 2.3.7.4.2.2

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

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

Step 2.3.7.4.2.3

Move the negative in front of the fraction.

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

Step 2.3.8

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

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

Step 2.3.9

Simplify.

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Step 2.3.9.1

Rewrite $- (2 {u_{2}}^{\frac{1}{2}} + C_{2})$ as $- 1 \cdot 2 {u_{2}}^{\frac{1}{2}} + C_{2}$ .

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

Step 2.3.9.2

Multiply $- 1$ by $2$ .

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

Step 2.3.10

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Solve for $y$ .

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Step 3.3.1

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

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

Step 3.3.2

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

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

Step 3.3.3

Subtract $2$ from both sides of the equation.

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

Step 4

Group the constant terms together.

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Step 4.1

Rewrite $e^{- 2 {(1 + x^{2})}^{\frac{1}{2}} + C}$ as $e^{- 2 {(1 + x^{2})}^{\frac{1}{2}}} e^{C}$ .

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

Step 4.2

Reorder $e^{- 2 {(1 + x^{2})}^{\frac{1}{2}}}$ and $e^{C}$ .

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

Step 4.3

Combine constants with the plus or minus.

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