Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Regroup factors.

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

Step 1.2

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

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

Step 1.3

Cancel the common factor of $e^{- y}$ .

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Step 1.3.1

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

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

Step 1.3.2

Cancel the common factor.

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

Step 1.3.3

Rewrite the expression.

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

Step 1.4

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Simplify the expression.

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Step 2.2.1.1

Negate the exponent of $e^{- y}$ and move it out of the denominator.

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

Step 2.2.1.2

Simplify.

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Step 2.2.1.2.1

Multiply the exponents in ${(e^{- y})}^{- 1}$ .

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Step 2.2.1.2.1.1

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

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

Step 2.2.1.2.1.2

Multiply $- y \cdot - 1$ .

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Step 2.2.1.2.1.2.1

Multiply $- 1$ by $- 1$ .

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

Step 2.2.1.2.1.2.2

Multiply $y$ by $1$ .

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

Step 2.2.1.2.2

Multiply $e^{y}$ by $1$ .

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

Step 2.2.2

The integral of $e^{y}$ with respect to $y$ is $e^{y}$ .

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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 2.3.1.1

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

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Step 2.3.1.1.1

Differentiate $1 + x^{2}$ .

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

Step 2.3.1.1.2

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 2.3.1.1.3

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 2.3.1.1.4

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 2.3.1.1.5

Add $0$ and $2 x$ .

$2 x$

Step 2.3.1.2

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

$e^{y} + C_{1} = \int \frac{1}{u} \cdot \frac{1}{2} d u$

Step 2.3.2

Simplify.

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Step 2.3.2.1

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

$e^{y} + C_{1} = \int \frac{1}{u \cdot 2} d u$

Step 2.3.2.2

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

$e^{y} + C_{1} = \int \frac{1}{2 u} d u$

Step 2.3.3

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

$e^{y} + C_{1} = \frac{1}{2} \int \frac{1}{u} d u$

Step 2.3.4

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

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

Step 2.3.5

Simplify.

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

Step 2.3.6

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

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

Step 3.2

Expand the left side.

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Step 3.2.1

Expand $\ln (e^{y})$ by moving $y$ outside the logarithm.

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

Step 3.2.2

The natural logarithm of $e$ is $1$ .

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

Step 3.2.3

Multiply $y$ by $1$ .

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

Step 3.3

Expand $\ln ({| 1 + x^{2} |}^{\frac{1}{2}})$ by moving $\frac{1}{2}$ outside the logarithm.

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

Step 3.4

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

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