Calculus Examples

$(4 + e^{2 y}) d x = x e^{2 y} d y$

Step 1

Rewrite the equation.

$x e^{2 y} d y = (4 + e^{2 y}) d x$

Step 2

Multiply both sides by $\frac{1}{x (4 + e^{2 y})}$ .

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

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $x$ .

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Step 3.1.1

Factor $x$ out of $x e^{2 y}$ .

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

Step 3.1.2

Cancel the common factor.

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

Step 3.1.3

Rewrite the expression.

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

Step 3.2

Combine $\frac{1}{4 + e^{2 y}}$ and $e^{2 y}$ .

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

Step 3.3

Cancel the common factor of $4 + e^{2 y}$ .

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Step 3.3.1

Factor $4 + e^{2 y}$ out of $x (4 + e^{2 y})$ .

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

Step 3.3.2

Cancel the common factor.

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

Step 3.3.3

Rewrite the expression.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

$\int \frac{e^{2 y}}{4 + e^{2 y}} d y = \int \frac{1}{x} d x$

Step 4.2

Integrate the left side.

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Step 4.2.1

Let $u_{2} = 4 + e^{2 y}$ . Then $d u_{2} = 2 e^{2 y} d y$ , so $\frac{1}{2} d u_{2} = e^{2 y} d y$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 4.2.1.1

Let $u_{2} = 4 + e^{2 y}$ . Find $\frac{d u_{2}}{d y}$ .

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Step 4.2.1.1.1

Differentiate $4 + e^{2 y}$ .

$\frac{d}{d y} [4 + e^{2 y}]$

Step 4.2.1.1.2

Differentiate.

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Step 4.2.1.1.2.1

By the Sum Rule, the derivative of $4 + e^{2 y}$ with respect to $y$ is $\frac{d}{d y} [4] + \frac{d}{d y} [e^{2 y}]$ .

$\frac{d}{d y} [4] + \frac{d}{d y} [e^{2 y}]$

Step 4.2.1.1.2.2

Since $4$ is constant with respect to $y$ , the derivative of $4$ with respect to $y$ is $0$ .

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

Step 4.2.1.1.3

Evaluate $\frac{d}{d y} [e^{2 y}]$ .

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Step 4.2.1.1.3.1

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = e^{y}$ and $g (y) = 2 y$ .

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Step 4.2.1.1.3.1.1

To apply the Chain Rule, set $u_{1}$ as $2 y$ .

$0 + \frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d y} [2 y]$

Step 4.2.1.1.3.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$0 + e^{u_{1}} \frac{d}{d y} [2 y]$

Step 4.2.1.1.3.1.3

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

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

Step 4.2.1.1.3.2

Since $2$ is constant with respect to $y$ , the derivative of $2 y$ with respect to $y$ is $2 \frac{d}{d y} [y]$ .

$0 + e^{2 y} (2 \frac{d}{d y} [y])$

Step 4.2.1.1.3.3

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

$0 + e^{2 y} (2 \cdot 1)$

Step 4.2.1.1.3.4

Multiply $2$ by $1$ .

$0 + e^{2 y} \cdot 2$

Step 4.2.1.1.3.5

Move $2$ to the left of $e^{2 y}$ .

$0 + 2 e^{2 y}$

Step 4.2.1.1.4

Add $0$ and $2 e^{2 y}$ .

$2 e^{2 y}$

Step 4.2.1.2

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

$\int \frac{1}{u_{2}} \cdot \frac{1}{2} d u_{2} = \int \frac{1}{x} d x$

Step 4.2.2

Simplify.

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Step 4.2.2.1

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

$\int \frac{1}{u_{2} \cdot 2} d u_{2} = \int \frac{1}{x} d x$

Step 4.2.2.2

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

$\int \frac{1}{2 u_{2}} d u_{2} = \int \frac{1}{x} d x$

Step 4.2.3

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

$\frac{1}{2} \int \frac{1}{u_{2}} d u_{2} = \int \frac{1}{x} d x$

Step 4.2.4

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

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

Step 4.2.5

Simplify.

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

Step 4.2.6

Replace all occurrences of $u_{2}$ with $4 + e^{2 y}$ .

$\frac{1}{2} \ln (| 4 + e^{2 y} |) + C_{1} = \int \frac{1}{x} d x$

Step 4.3

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

$\frac{1}{2} \ln (| 4 + e^{2 y} |) + C_{1} = \ln (| x |) + C_{2}$

Step 4.4

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

$\frac{1}{2} \ln (| 4 + e^{2 y} |) = \ln (| x |) + C$