Calculus Examples

$(x^{2} + 1) \frac{d y}{d x} + 3 x y = 6 x$

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

Simplify each term.

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Step 1.1.1.1

Apply the distributive property.

$x^{2} \frac{d y}{d x} + 1 \frac{d y}{d x} + 3 x y = 6 x$

Step 1.1.1.2

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

$x^{2} \frac{d y}{d x} + \frac{d y}{d x} + 3 x y = 6 x$

Step 1.1.2

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

$x^{2} \frac{d y}{d x} + \frac{d y}{d x} = 6 x - 3 x y$

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 $x^{2} \frac{d y}{d x}$ .

$\frac{d y}{d x} x^{2} + \frac{d y}{d x} = 6 x - 3 x y$

Step 1.1.3.2

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

$\frac{d y}{d x} x^{2} + \frac{d y}{d x} = 6 x - 3 x y$

Step 1.1.3.3

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

$\frac{d y}{d x} x^{2} + \frac{d y}{d x} \cdot 1 = 6 x - 3 x y$

Step 1.1.3.4

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

$\frac{d y}{d x} (x^{2} + 1) = 6 x - 3 x y$

Step 1.1.4

Divide each term in $\frac{d y}{d x} (x^{2} + 1) = 6 x - 3 x y$ by $x^{2} + 1$ and simplify.

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Step 1.1.4.1

Divide each term in $\frac{d y}{d x} (x^{2} + 1) = 6 x - 3 x y$ by $x^{2} + 1$ .

$\frac{\frac{d y}{d x} (x^{2} + 1)}{x^{2} + 1} = \frac{6 x}{x^{2} + 1} + \frac{- 3 x y}{x^{2} + 1}$

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.

$\frac{\frac{d y}{d x} (x^{2} + 1)}{x^{2} + 1} = \frac{6 x}{x^{2} + 1} + \frac{- 3 x y}{x^{2} + 1}$

Step 1.1.4.2.1.2

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

$\frac{d y}{d x} = \frac{6 x}{x^{2} + 1} + \frac{- 3 x y}{x^{2} + 1}$

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.

$\frac{d y}{d x} = \frac{6 x - 3 x y}{x^{2} + 1}$

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 $6 x - 3 x y$ .

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Step 1.1.4.3.2.1.1

Factor $3 x$ out of $6 x$ .

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

Step 1.1.4.3.2.1.2

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

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

Step 1.1.4.3.2.1.3

Factor $3 x$ out of $3 x (2) + 3 x (- 1 y)$ .

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

Step 1.1.4.3.2.2

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

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

Step 1.2

Regroup factors.

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

Step 1.3

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

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

Step 1.4

Cancel the common factor of $2 - y$ .

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Step 1.4.1

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

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

Step 1.4.2

Cancel the common factor.

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

Step 1.4.3

Rewrite the expression.

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u_{1} = 2 - y$ . Then $d u_{1} = - d y$ , so $- d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Rewrite.

$\frac{- 1}{1}$

Step 2.2.1.1.2

Divide $- 1$ by $1$ .

$- 1$

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

Split the fraction into multiple fractions.

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

Step 2.2.3

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

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

Step 2.2.4

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 2.2.5

Simplify.

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

Step 2.2.6

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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

Step 2.3.2

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 2.3.2.1

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

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Step 2.3.2.1.1

Differentiate $x^{2} + 1$ .

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

Step 2.3.2.1.2

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 2.3.2.1.3

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 2.3.2.1.4

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

$2 x + 0$

Step 2.3.2.1.5

Add $2 x$ and $0$ .

$2 x$

Step 2.3.2.2

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

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

Step 2.3.3

Simplify.

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Step 2.3.3.1

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

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

Step 2.3.3.2

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

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

Step 2.3.4

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

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

Step 2.3.5

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

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

Step 2.3.6

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

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

Step 2.3.7

Simplify.

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

Step 2.3.8

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Simplify the right side.

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Step 3.1.1

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

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

Combine the numerators over the common denominator.

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

Step 3.6

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

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Step 3.6.1

Reorder the expression.

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Step 3.6.1.1

Reorder $2$ and $- y$ .

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

Step 3.6.1.2

Move $\ln (| - y + 2 |)$ .

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

Step 3.6.2

Factor $- 1$ out of $- 1 \cdot 2 \ln (| - y + 2 |)$ .

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

Step 3.6.3

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

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

Step 3.6.4

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

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

Step 3.7

Move the negative in front of the fraction.

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

Step 3.8

Simplify the left side.

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Step 3.8.1

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

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Step 3.8.1.1

Simplify the numerator.

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Step 3.8.1.1.1

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

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

Step 3.8.1.1.2

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

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

Step 3.8.1.1.3

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

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

Step 3.8.1.1.4

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

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

Step 3.8.1.2

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

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

Step 3.8.1.3

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

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

Step 3.8.1.4

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

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

Step 3.8.1.5

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

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Step 3.8.1.5.1

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

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

Step 3.8.1.5.2

Cancel the common factor of $2$ .

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Step 3.8.1.5.2.1

Cancel the common factor.

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

Step 3.8.1.5.2.2

Rewrite the expression.

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

Step 3.8.1.6

Simplify.

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

Step 3.8.1.7

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

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Step 3.8.1.7.1

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

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

Step 3.8.1.7.2

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

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

Step 3.8.1.8

Apply the distributive property.

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

Step 3.9

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

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Step 3.9.1

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

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

Step 3.9.2

Simplify the left side.

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Step 3.9.2.1

Dividing two negative values results in a positive value.

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

Step 3.9.2.2

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

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

Step 3.9.3

Simplify the right side.

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Step 3.9.3.1

Move the negative one from the denominator of $\frac{C}{- 1}$ .

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

Step 3.9.3.2

Rewrite $- 1 \cdot C$ as $- C$ .

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

Step 3.10

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

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

Step 3.11

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

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

Step 3.12

Solve for $y$ .

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Step 3.12.1

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

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

Step 3.12.2

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

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

Step 3.12.3

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

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Step 3.12.3.1

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

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

Step 3.12.3.2

Simplify the left side.

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Step 3.12.3.2.1

Dividing two negative values results in a positive value.

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

Step 3.12.3.2.2

Cancel the common factor.

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

Step 3.12.3.2.3

Divide $y$ by $1$ .

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

Step 3.12.3.3

Simplify the right side.

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Step 3.12.3.3.1

To write $\frac{e^{- C}}{- {| x^{2} + 1 |}^{\frac{3}{2}}}$ as a fraction with a common denominator, multiply by $\frac{- 1}{- 1}$ .

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

Step 3.12.3.3.2

To write $\frac{- 2 {| x^{2} + 1 |}^{\frac{3}{2}}}{- {| x^{2} + 1 |}^{\frac{3}{2}}}$ as a fraction with a common denominator, multiply by $\frac{- 1}{- 1}$ .

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

Step 3.12.3.3.3

Write each expression with a common denominator of ${| x^{2} + 1 |}^{\frac{3}{2}}$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.12.3.3.3.1

Multiply $\frac{e^{- C}}{- {| x^{2} + 1 |}^{\frac{3}{2}}}$ by $\frac{- 1}{- 1}$ .

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

Step 3.12.3.3.3.2

Multiply $- 1$ by $- 1$ .

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

Step 3.12.3.3.3.3

Multiply ${| x^{2} + 1 |}^{\frac{3}{2}}$ by $1$ .

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

Step 3.12.3.3.3.4

Multiply $\frac{- 2 {| x^{2} + 1 |}^{\frac{3}{2}}}{- {| x^{2} + 1 |}^{\frac{3}{2}}}$ by $\frac{- 1}{- 1}$ .

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

Step 3.12.3.3.3.5

Multiply $- 1$ by $- 1$ .

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

Step 3.12.3.3.3.6

Multiply ${| x^{2} + 1 |}^{\frac{3}{2}}$ by $1$ .

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

Step 3.12.3.3.4

Combine the numerators over the common denominator.

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

Step 3.12.3.3.5

Simplify each term.

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Step 3.12.3.3.5.1

Move $- 1$ to the left of $e^{- C}$ .

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

Step 3.12.3.3.5.2

Rewrite $- 1 e^{- C}$ as $- e^{- C}$ .

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

Step 3.12.3.3.5.3

Multiply $- 1$ by $- 2$ .

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

Step 3.12.3.3.6

Simplify with factoring out.

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Step 3.12.3.3.6.1

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

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

Step 3.12.3.3.6.2

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

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

Step 3.12.3.3.6.3

Simplify the expression.

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Step 3.12.3.3.6.3.1

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

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

Step 3.12.3.3.6.3.2

Move the negative in front of the fraction.

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

Step 4

Simplify the constant of integration.

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