Calculus Examples

$(x^{2} + 1) \frac{d y}{d x} + 3 x (y - 1) = 0$ , $y (0) = 9$

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

Simplify each term.

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Step 1.1.1.1

Apply the distributive property.

$x^{2} \frac{d y}{d x} + 1 \frac{d y}{d x} + 3 x (y - 1) = 0$

Step 1.1.1.2

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

$x^{2} \frac{d y}{d x} + \frac{d y}{d x} + 3 x (y - 1) = 0$

Step 1.1.1.3

Apply the distributive property.

$x^{2} \frac{d y}{d x} + \frac{d y}{d x} + 3 x y + 3 x \cdot - 1 = 0$

Step 1.1.1.4

Multiply $- 1$ by $3$ .

$x^{2} \frac{d y}{d x} + \frac{d y}{d x} + 3 x y - 3 x = 0$

Step 1.1.2

Move all terms not containing $\frac{d y}{d x}$ to the right side of the equation.

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Step 1.1.2.1

Subtract $3 x y$ from both sides of the equation.

$x^{2} \frac{d y}{d x} + \frac{d y}{d x} - 3 x = - 3 x y$

Step 1.1.2.2

Add $3 x$ to both sides of the equation.

$x^{2} \frac{d y}{d x} + \frac{d y}{d x} = - 3 x y + 3 x$

Step 1.1.3

Factor $\frac{d y}{d x}$ out of $x^{2} \frac{d y}{d x} + \frac{d y}{d x}$ .

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Step 1.1.3.1

Factor $\frac{d y}{d x}$ out of $x^{2} \frac{d y}{d x}$ .

$\frac{d y}{d x} x^{2} + \frac{d y}{d x} = - 3 x y + 3 x$

Step 1.1.3.2

Raise $\frac{d y}{d x}$ to the power of $1$ .

$\frac{d y}{d x} x^{2} + \frac{d y}{d x} = - 3 x y + 3 x$

Step 1.1.3.3

Factor $\frac{d y}{d x}$ out of ${(\frac{d y}{d x})}^{1}$ .

$\frac{d y}{d x} x^{2} + \frac{d y}{d x} \cdot 1 = - 3 x y + 3 x$

Step 1.1.3.4

Factor $\frac{d y}{d x}$ out of $\frac{d y}{d x} x^{2} + \frac{d y}{d x} \cdot 1$ .

$\frac{d y}{d x} (x^{2} + 1) = - 3 x y + 3 x$

Step 1.1.4

Divide each term in $\frac{d y}{d x} (x^{2} + 1) = - 3 x y + 3 x$ by $x^{2} + 1$ and simplify.

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Step 1.1.4.1

Divide each term in $\frac{d y}{d x} (x^{2} + 1) = - 3 x y + 3 x$ by $x^{2} + 1$ .

$\frac{\frac{d y}{d x} (x^{2} + 1)}{x^{2} + 1} = \frac{- 3 x y}{x^{2} + 1} + \frac{3 x}{x^{2} + 1}$

Step 1.1.4.2

Simplify the left side.

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Step 1.1.4.2.1

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

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Step 1.1.4.2.1.1

Cancel the common factor.

$\frac{\frac{d y}{d x} (x^{2} + 1)}{x^{2} + 1} = \frac{- 3 x y}{x^{2} + 1} + \frac{3 x}{x^{2} + 1}$

Step 1.1.4.2.1.2

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

$\frac{d y}{d x} = \frac{- 3 x y}{x^{2} + 1} + \frac{3 x}{x^{2} + 1}$

Step 1.1.4.3

Simplify the right side.

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Step 1.1.4.3.1

Combine the numerators over the common denominator.

$\frac{d y}{d x} = \frac{- 3 x y + 3 x}{x^{2} + 1}$

Step 1.1.4.3.2

Simplify the numerator.

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Step 1.1.4.3.2.1

Factor $3 x$ out of $- 3 x y + 3 x$ .

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Step 1.1.4.3.2.1.1

Factor $3 x$ out of $- 3 x y$ .

$\frac{d y}{d x} = \frac{3 x (- 1 y) + 3 x}{x^{2} + 1}$

Step 1.1.4.3.2.1.2

Factor $3 x$ out of $3 x$ .

$\frac{d y}{d x} = \frac{3 x (- 1 y) + 3 x (1)}{x^{2} + 1}$

Step 1.1.4.3.2.1.3

Factor $3 x$ out of $3 x (- 1 y) + 3 x (1)$ .

$\frac{d y}{d x} = \frac{3 x (- 1 y + 1)}{x^{2} + 1}$

Step 1.1.4.3.2.2

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

$\frac{d y}{d x} = \frac{3 x (- y + 1)}{x^{2} + 1}$

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 $- y$ .

$\frac{d y}{d x} = \frac{3 x (- (y) + 1)}{x^{2} + 1}$

Step 1.1.4.3.3.2

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

$\frac{d y}{d x} = \frac{3 x (- (y) - 1 (- 1))}{x^{2} + 1}$

Step 1.1.4.3.3.3

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

$\frac{d y}{d x} = \frac{3 x (- (y - 1))}{x^{2} + 1}$

Step 1.1.4.3.3.4

Simplify the expression.

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Step 1.1.4.3.3.4.1

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

$\frac{d y}{d x} = \frac{3 x (- 1 (y - 1))}{x^{2} + 1}$

Step 1.1.4.3.3.4.2

Move the negative in front of the fraction.

$\frac{d y}{d x} = - \frac{(3 x) (y - 1)}{x^{2} + 1}$

Step 1.2

Regroup factors.

$\frac{d y}{d x} = - \frac{3 x}{x^{2} + 1} (y - 1)$

Step 1.3

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

$\frac{1}{y - 1} \frac{d y}{d x} = \frac{1}{y - 1} (- \frac{3 x}{x^{2} + 1} (y - 1))$

Step 1.4

Simplify.

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Step 1.4.1

Rewrite using the commutative property of multiplication.

$\frac{1}{y - 1} \frac{d y}{d x} = - \frac{1}{y - 1} (\frac{3 x}{x^{2} + 1} (y - 1))$

Step 1.4.2

Cancel the common factor of $y - 1$ .

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Step 1.4.2.1

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

$\frac{1}{y - 1} \frac{d y}{d x} = \frac{- 1}{y - 1} (\frac{3 x}{x^{2} + 1} (y - 1))$

Step 1.4.2.2

Factor $y - 1$ out of $\frac{3 x}{x^{2} + 1} (y - 1)$ .

$\frac{1}{y - 1} \frac{d y}{d x} = \frac{- 1}{y - 1} ((y - 1) \frac{3 x}{x^{2} + 1})$

Step 1.4.2.3

Cancel the common factor.

$\frac{1}{y - 1} \frac{d y}{d x} = \frac{- 1}{y - 1} ((y - 1) \frac{3 x}{x^{2} + 1})$

Step 1.4.2.4

Rewrite the expression.

$\frac{1}{y - 1} \frac{d y}{d x} = - \frac{3 x}{x^{2} + 1}$

Step 1.5

Rewrite the equation.

$\frac{1}{y - 1} d y = - \frac{3 x}{x^{2} + 1} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y - 1} d y = \int - \frac{3 x}{x^{2} + 1} d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Differentiate $y - 1$ .

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

Step 2.2.1.1.2

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 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 $n = 1$ .

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

Step 2.2.1.1.4

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

$1 + 0$

Step 2.2.1.1.5

Add $1$ and $0$ .

$1$

Step 2.2.1.2

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

$\int \frac{1}{u_{1}} d u_{1} = \int - \frac{3 x}{x^{2} + 1} d x$

Step 2.2.2

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

$\ln (| u_{1} |) + C_{1} = \int - \frac{3 x}{x^{2} + 1} d x$

Step 2.2.3

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

$\ln (| y - 1 |) + C_{1} = \int - \frac{3 x}{x^{2} + 1} d x$

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.

$\ln (| y - 1 |) + C_{1} = - \int \frac{3 x}{x^{2} + 1} d x$

Step 2.3.2

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

$\ln (| y - 1 |) + C_{1} = - (3 \int \frac{x}{x^{2} + 1} d x)$

Step 2.3.3

Multiply $3$ by $- 1$ .

$\ln (| y - 1 |) + C_{1} = - 3 \int \frac{x}{x^{2} + 1} d x$

Step 2.3.4

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

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Step 2.3.4.1.1

Differentiate $x^{2} + 1$ .

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

Step 2.3.4.1.2

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 2.3.4.1.3

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 2.3.4.1.4

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

$2 x + 0$

Step 2.3.4.1.5

Add $2 x$ and $0$ .

$2 x$

Step 2.3.4.2

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

$\ln (| y - 1 |) + C_{1} = - 3 \int \frac{1}{u_{2}} \cdot \frac{1}{2} d u_{2}$

Step 2.3.5

Simplify.

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Step 2.3.5.1

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

$\ln (| y - 1 |) + C_{1} = - 3 \int \frac{1}{u_{2} \cdot 2} d u_{2}$

Step 2.3.5.2

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

$\ln (| y - 1 |) + C_{1} = - 3 \int \frac{1}{2 u_{2}} d u_{2}$

Step 2.3.6

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

$\ln (| y - 1 |) + C_{1} = - 3 (\frac{1}{2} \int \frac{1}{u_{2}} d u_{2})$

Step 2.3.7

Simplify.

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Step 2.3.7.1

Combine $\frac{1}{2}$ and $- 3$ .

$\ln (| y - 1 |) + C_{1} = \frac{- 3}{2} \int \frac{1}{u_{2}} d u_{2}$

Step 2.3.7.2

Move the negative in front of the fraction.

$\ln (| y - 1 |) + C_{1} = - \frac{3}{2} \int \frac{1}{u_{2}} d u_{2}$

Step 2.3.8

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

$\ln (| y - 1 |) + C_{1} = - \frac{3}{2} (\ln (| u_{2} |) + C_{2})$

Step 2.3.9

Simplify.

$\ln (| y - 1 |) + C_{1} = - \frac{3}{2} \ln (| u_{2} |) + C_{2}$

Step 2.3.10

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

$\ln (| y - 1 |) + C_{1} = - \frac{3}{2} \ln (| x^{2} + 1 |) + C_{2}$

Step 2.4

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

$\ln (| y - 1 |) = - \frac{3}{2} \ln (| x^{2} + 1 |) + C$

Step 3

Solve for $y$ .

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Step 3.1

Simplify the right side.

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Step 3.1.1

Simplify each term.

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Step 3.1.1.1

Combine $\ln (| x^{2} + 1 |)$ and $\frac{3}{2}$ .

$\ln (| y - 1 |) = - \frac{\ln (| x^{2} + 1 |) \cdot 3}{2} + C$

Step 3.1.1.2

Move $3$ to the left of $\ln (| x^{2} + 1 |)$ .

$\ln (| y - 1 |) = - \frac{3 \ln (| x^{2} + 1 |)}{2} + C$

Step 3.2

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

$\ln (| y - 1 |) + \frac{3 \ln (| x^{2} + 1 |)}{2} = C$

Step 3.3

To write $\ln (| y - 1 |)$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\ln (| y - 1 |) \cdot \frac{2}{2} + \frac{3 \ln (| x^{2} + 1 |)}{2} = C$

Step 3.4

Simplify terms.

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Step 3.4.1

Combine $\ln (| y - 1 |)$ and $\frac{2}{2}$ .

$\frac{\ln (| y - 1 |) \cdot 2}{2} + \frac{3 \ln (| x^{2} + 1 |)}{2} = C$

Step 3.4.2

Combine the numerators over the common denominator.

$\frac{\ln (| y - 1 |) \cdot 2 + 3 \ln (| x^{2} + 1 |)}{2} = C$

Step 3.5

Move $2$ to the left of $\ln (| y - 1 |)$ .

$\frac{2 \ln (| y - 1 |) + 3 \ln (| x^{2} + 1 |)}{2} = C$

Step 3.6

Simplify the left side.

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Step 3.6.1

Simplify $\frac{2 \ln (| y - 1 |) + 3 \ln (| x^{2} + 1 |)}{2}$ .

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Step 3.6.1.1

Simplify the numerator.

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Step 3.6.1.1.1

Simplify $2 \ln (| y - 1 |)$ by moving $2$ inside the logarithm.

$\frac{\ln ({| y - 1 |}^{2}) + 3 \ln (| x^{2} + 1 |)}{2} = C$

Step 3.6.1.1.2

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

$\frac{\ln ({(y - 1)}^{2}) + 3 \ln (| x^{2} + 1 |)}{2} = C$

Step 3.6.1.1.3

Simplify $3 \ln (| x^{2} + 1 |)$ by moving $3$ inside the logarithm.

$\frac{\ln ({(y - 1)}^{2}) + \ln ({| x^{2} + 1 |}^{3})}{2} = C$

Step 3.6.1.1.4

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

$\frac{\ln ({(y - 1)}^{2} {| x^{2} + 1 |}^{3})}{2} = C$

Step 3.6.1.2

Rewrite $\frac{\ln ({(y - 1)}^{2} {| x^{2} + 1 |}^{3})}{2}$ as $\frac{1}{2} \ln ({(y - 1)}^{2} {| x^{2} + 1 |}^{3})$ .

$\frac{1}{2} \ln ({(y - 1)}^{2} {| x^{2} + 1 |}^{3}) = C$

Step 3.6.1.3

Simplify $\frac{1}{2} \ln ({(y - 1)}^{2} {| x^{2} + 1 |}^{3})$ by moving $\frac{1}{2}$ inside the logarithm.

$\ln ({({(y - 1)}^{2} {| x^{2} + 1 |}^{3})}^{\frac{1}{2}}) = C$

Step 3.6.1.4

Apply the product rule to ${(y - 1)}^{2} {| x^{2} + 1 |}^{3}$ .

$\ln ({({(y - 1)}^{2})}^{\frac{1}{2}} {({| x^{2} + 1 |}^{3})}^{\frac{1}{2}}) = C$

Step 3.6.1.5

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

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Step 3.6.1.5.1

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

$\ln ({(y - 1)}^{2 (\frac{1}{2})} {({| x^{2} + 1 |}^{3})}^{\frac{1}{2}}) = C$

Step 3.6.1.5.2

Cancel the common factor of $2$ .

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Step 3.6.1.5.2.1

Cancel the common factor.

$\ln ({(y - 1)}^{2 (\frac{1}{2})} {({| x^{2} + 1 |}^{3})}^{\frac{1}{2}}) = C$

Step 3.6.1.5.2.2

Rewrite the expression.

$\ln ({(y - 1)}^{1} {({| x^{2} + 1 |}^{3})}^{\frac{1}{2}}) = C$

Step 3.6.1.6

Simplify.

$\ln ((y - 1) {({| x^{2} + 1 |}^{3})}^{\frac{1}{2}}) = C$

Step 3.6.1.7

Multiply the exponents in ${({| x^{2} + 1 |}^{3})}^{\frac{1}{2}}$ .

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Step 3.6.1.7.1

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

$\ln ((y - 1) {| x^{2} + 1 |}^{3 (\frac{1}{2})}) = C$

Step 3.6.1.7.2

Combine $3$ and $\frac{1}{2}$ .

$\ln ((y - 1) {| x^{2} + 1 |}^{\frac{3}{2}}) = C$

Step 3.6.1.8

Apply the distributive property.

$\ln (y {| x^{2} + 1 |}^{\frac{3}{2}} - 1 {| x^{2} + 1 |}^{\frac{3}{2}}) = C$

Step 3.6.1.9

Rewrite $- 1 {| x^{2} + 1 |}^{\frac{3}{2}}$ as $- {| x^{2} + 1 |}^{\frac{3}{2}}$ .

$\ln (y {| x^{2} + 1 |}^{\frac{3}{2}} - {| x^{2} + 1 |}^{\frac{3}{2}}) = C$

Step 3.7

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

$e^{\ln (y {| x^{2} + 1 |}^{\frac{3}{2}} - {| x^{2} + 1 |}^{\frac{3}{2}})} = e^{C}$

Step 3.8

Rewrite $\ln (y {| x^{2} + 1 |}^{\frac{3}{2}} - {| x^{2} + 1 |}^{\frac{3}{2}}) = C$ 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 $b^{y} = x$ .

$e^{C} = y {| x^{2} + 1 |}^{\frac{3}{2}} - {| x^{2} + 1 |}^{\frac{3}{2}}$

Step 3.9

Solve for $y$ .

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Step 3.9.1

Rewrite the equation as $y {| x^{2} + 1 |}^{\frac{3}{2}} - {| x^{2} + 1 |}^{\frac{3}{2}} = e^{C}$ .

$y {| x^{2} + 1 |}^{\frac{3}{2}} - {| x^{2} + 1 |}^{\frac{3}{2}} = e^{C}$

Step 3.9.2

Add ${| x^{2} + 1 |}^{\frac{3}{2}}$ to both sides of the equation.

$y {| x^{2} + 1 |}^{\frac{3}{2}} = e^{C} + {| x^{2} + 1 |}^{\frac{3}{2}}$

Step 3.9.3

Divide each term in $y {| x^{2} + 1 |}^{\frac{3}{2}} = e^{C} + {| x^{2} + 1 |}^{\frac{3}{2}}$ by ${| x^{2} + 1 |}^{\frac{3}{2}}$ and simplify.

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Step 3.9.3.1

Divide each term in $y {| x^{2} + 1 |}^{\frac{3}{2}} = e^{C} + {| x^{2} + 1 |}^{\frac{3}{2}}$ by ${| x^{2} + 1 |}^{\frac{3}{2}}$ .

$\frac{y {| x^{2} + 1 |}^{\frac{3}{2}}}{{| x^{2} + 1 |}^{\frac{3}{2}}} = \frac{e^{C}}{{| x^{2} + 1 |}^{\frac{3}{2}}} + \frac{{| x^{2} + 1 |}^{\frac{3}{2}}}{{| x^{2} + 1 |}^{\frac{3}{2}}}$

Step 3.9.3.2

Simplify the left side.

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Step 3.9.3.2.1

Cancel the common factor.

$\frac{y {| x^{2} + 1 |}^{\frac{3}{2}}}{{| x^{2} + 1 |}^{\frac{3}{2}}} = \frac{e^{C}}{{| x^{2} + 1 |}^{\frac{3}{2}}} + \frac{{| x^{2} + 1 |}^{\frac{3}{2}}}{{| x^{2} + 1 |}^{\frac{3}{2}}}$

Step 3.9.3.2.2

Divide $y$ by $1$ .

$y = \frac{e^{C}}{{| x^{2} + 1 |}^{\frac{3}{2}}} + \frac{{| x^{2} + 1 |}^{\frac{3}{2}}}{{| x^{2} + 1 |}^{\frac{3}{2}}}$

Step 3.9.3.3

Simplify the right side.

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Step 3.9.3.3.1

Combine the numerators over the common denominator.

$y = \frac{e^{C} + {| x^{2} + 1 |}^{\frac{3}{2}}}{{| x^{2} + 1 |}^{\frac{3}{2}}}$

Step 4

Simplify the constant of integration.

$y = \frac{C + {| x^{2} + 1 |}^{\frac{3}{2}}}{{| x^{2} + 1 |}^{\frac{3}{2}}}$

Step 5

Use the initial condition to find the value of $C$ by substituting $0$ for $x$ and $9$ for $y$ in $y = \frac{C + {| x^{2} + 1 |}^{\frac{3}{2}}}{{| x^{2} + 1 |}^{\frac{3}{2}}}$ .

$9 = \frac{C + {| 0^{2} + 1 |}^{\frac{3}{2}}}{{| 0^{2} + 1 |}^{\frac{3}{2}}}$

Step 6

Solve for $C$ .

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Step 6.1

Rewrite the equation as $\frac{C + {| 0^{2} + 1 |}^{\frac{3}{2}}}{{| 0^{2} + 1 |}^{\frac{3}{2}}} = 9$ .

$\frac{C + {| 0^{2} + 1 |}^{\frac{3}{2}}}{{| 0^{2} + 1 |}^{\frac{3}{2}}} = 9$

Step 6.2

Multiply both sides of the equation by ${| 0^{2} + 1 |}^{\frac{3}{2}}$ .

${| 0^{2} + 1 |}^{\frac{3}{2}} \frac{C + {| 0^{2} + 1 |}^{\frac{3}{2}}}{{| 0^{2} + 1 |}^{\frac{3}{2}}} = {| 0^{2} + 1 |}^{\frac{3}{2}} \cdot 9$

Step 6.3

Simplify both sides of the equation.

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Step 6.3.1

Simplify the left side.

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Step 6.3.1.1

Simplify ${| 0^{2} + 1 |}^{\frac{3}{2}} \frac{C + {| 0^{2} + 1 |}^{\frac{3}{2}}}{{| 0^{2} + 1 |}^{\frac{3}{2}}}$ .

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Step 6.3.1.1.1

Cancel the common factor of ${| 0^{2} + 1 |}^{\frac{3}{2}}$ .

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Step 6.3.1.1.1.1

Cancel the common factor.

${| 0^{2} + 1 |}^{\frac{3}{2}} \frac{C + {| 0^{2} + 1 |}^{\frac{3}{2}}}{{| 0^{2} + 1 |}^{\frac{3}{2}}} = {| 0^{2} + 1 |}^{\frac{3}{2}} \cdot 9$

Step 6.3.1.1.1.2

Rewrite the expression.

$C + {| 0^{2} + 1 |}^{\frac{3}{2}} = {| 0^{2} + 1 |}^{\frac{3}{2}} \cdot 9$

Step 6.3.1.1.2

Simplify each term.

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Step 6.3.1.1.2.1

Raising $0$ to any positive power yields $0$ .

$C + {| 0 + 1 |}^{\frac{3}{2}} = {| 0^{2} + 1 |}^{\frac{3}{2}} \cdot 9$

Step 6.3.1.1.2.2

Add $0$ and $1$ .

$C + {| 1 |}^{\frac{3}{2}} = {| 0^{2} + 1 |}^{\frac{3}{2}} \cdot 9$

Step 6.3.1.1.2.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$C + 1^{\frac{3}{2}} = {| 0^{2} + 1 |}^{\frac{3}{2}} \cdot 9$

Step 6.3.1.1.2.4

One to any power is one.

$C + 1 = {| 0^{2} + 1 |}^{\frac{3}{2}} \cdot 9$

Step 6.3.2

Simplify the right side.

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Step 6.3.2.1

Simplify ${| 0^{2} + 1 |}^{\frac{3}{2}} \cdot 9$ .

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Step 6.3.2.1.1

Raising $0$ to any positive power yields $0$ .

$C + 1 = {| 0 + 1 |}^{\frac{3}{2}} \cdot 9$

Step 6.3.2.1.2

Add $0$ and $1$ .

$C + 1 = {| 1 |}^{\frac{3}{2}} \cdot 9$

Step 6.3.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$C + 1 = 1^{\frac{3}{2}} \cdot 9$

Step 6.3.2.1.4

One to any power is one.

$C + 1 = 1 \cdot 9$

Step 6.3.2.1.5

Multiply $9$ by $1$ .

$C + 1 = 9$

Step 6.4

Move all terms not containing $C$ to the right side of the equation.

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Step 6.4.1

Subtract $1$ from both sides of the equation.

$C = 9 - 1$

Step 6.4.2

Subtract $1$ from $9$ .

$C = 8$

Step 7

Substitute $8$ for $C$ in $y = \frac{C + {| x^{2} + 1 |}^{\frac{3}{2}}}{{| x^{2} + 1 |}^{\frac{3}{2}}}$ and simplify.

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Step 7.1

Substitute $8$ for $C$ .

$y = \frac{8 + {| x^{2} + 1 |}^{\frac{3}{2}}}{{| x^{2} + 1 |}^{\frac{3}{2}}}$

Step 7.2

Simplify the numerator.

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Step 7.2.1

Rewrite $8$ as $2^{3}$ .

$y = \frac{2^{3} + {| x^{2} + 1 |}^{\frac{3}{2}}}{{| x^{2} + 1 |}^{\frac{3}{2}}}$

Step 7.2.2

Rewrite ${| x^{2} + 1 |}^{\frac{3}{2}}$ as ${({| x^{2} + 1 |}^{\frac{1}{2}})}^{3}$ .

$y = \frac{2^{3} + {({| x^{2} + 1 |}^{\frac{1}{2}})}^{3}}{{| x^{2} + 1 |}^{\frac{3}{2}}}$

Step 7.2.3

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = 2$ and $b = {| x^{2} + 1 |}^{\frac{1}{2}}$ .

$y = \frac{(2 + {| x^{2} + 1 |}^{\frac{1}{2}}) (2^{2} - 2 {| x^{2} + 1 |}^{\frac{1}{2}} + {({| x^{2} + 1 |}^{\frac{1}{2}})}^{2})}{{| x^{2} + 1 |}^{\frac{3}{2}}}$

Step 7.2.4

Simplify.

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Step 7.2.4.1

Raise $2$ to the power of $2$ .

$y = \frac{(2 + {| x^{2} + 1 |}^{\frac{1}{2}}) (4 - 2 {| x^{2} + 1 |}^{\frac{1}{2}} + {({| x^{2} + 1 |}^{\frac{1}{2}})}^{2})}{{| x^{2} + 1 |}^{\frac{3}{2}}}$

Step 7.2.4.2

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

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Step 7.2.4.2.1

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

$y = \frac{(2 + {| x^{2} + 1 |}^{\frac{1}{2}}) (4 - 2 {| x^{2} + 1 |}^{\frac{1}{2}} + {| x^{2} + 1 |}^{\frac{1}{2} \cdot 2})}{{| x^{2} + 1 |}^{\frac{3}{2}}}$

Step 7.2.4.2.2

Cancel the common factor of $2$ .

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Step 7.2.4.2.2.1

Cancel the common factor.

$y = \frac{(2 + {| x^{2} + 1 |}^{\frac{1}{2}}) (4 - 2 {| x^{2} + 1 |}^{\frac{1}{2}} + {| x^{2} + 1 |}^{\frac{1}{2} \cdot 2})}{{| x^{2} + 1 |}^{\frac{3}{2}}}$

Step 7.2.4.2.2.2

Rewrite the expression.

$y = \frac{(2 + {| x^{2} + 1 |}^{\frac{1}{2}}) (4 - 2 {| x^{2} + 1 |}^{\frac{1}{2}} + {| x^{2} + 1 |}^{1})}{{| x^{2} + 1 |}^{\frac{3}{2}}}$

Step 7.2.4.3

Simplify.

$y = \frac{(2 + {| x^{2} + 1 |}^{\frac{1}{2}}) (4 - 2 {| x^{2} + 1 |}^{\frac{1}{2}} + | x^{2} + 1 |)}{{| x^{2} + 1 |}^{\frac{3}{2}}}$

Step 7.2.4.4

Reorder terms.

$y = \frac{(2 + {| x^{2} + 1 |}^{\frac{1}{2}}) (4 + | x^{2} + 1 | - 2 {| x^{2} + 1 |}^{\frac{1}{2}})}{{| x^{2} + 1 |}^{\frac{3}{2}}}$