Calculus Examples

$2 x e^{2 t} d t + (1 + e^{2 t}) d x = 0$

Step 1

Subtract $2 x e^{2 t} d t$ from both sides of the equation.

$(1 + e^{2 t}) d x = - 2 x e^{2 t} d t$

Step 2

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

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

Step 3

Simplify.

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Step 3.1

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

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Step 3.1.1

Factor $1 + e^{2 t}$ out of $(1 + e^{2 t}) x$ .

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

Step 3.1.2

Cancel the common factor.

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

Step 3.1.3

Rewrite the expression.

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

Step 3.2

Rewrite using the commutative property of multiplication.

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

Step 3.3

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

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

Step 3.4

Cancel the common factor of $x$ .

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Step 3.4.1

Factor $x$ out of $(1 + e^{2 t}) x$ .

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

Step 3.4.2

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

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

Step 3.4.3

Cancel the common factor.

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

Step 3.4.4

Rewrite the expression.

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

Step 3.5

Combine $\frac{- 2}{1 + e^{2 t}}$ and $e^{2 t}$ .

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

Step 3.6

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

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

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

Step 4.3

Integrate the right side.

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Step 4.3.1

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

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

Step 4.3.2

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

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

Step 4.3.3

Multiply $2$ by $- 1$ .

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

Step 4.3.4

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

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Step 4.3.4.1

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

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Step 4.3.4.1.1

Differentiate $1 + e^{2 t}$ .

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

Step 4.3.4.1.2

Differentiate.

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Step 4.3.4.1.2.1

By the Sum Rule, the derivative of $1 + e^{2 t}$ with respect to $t$ is $\frac{d}{d t} [1] + \frac{d}{d t} [e^{2 t}]$ .

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

Step 4.3.4.1.2.2

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

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

Step 4.3.4.1.3

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

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Step 4.3.4.1.3.1

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = 2 t$ .

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Step 4.3.4.1.3.1.1

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

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

Step 4.3.4.1.3.1.2

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

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

Step 4.3.4.1.3.1.3

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

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

Step 4.3.4.1.3.2

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

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

Step 4.3.4.1.3.3

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

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

Step 4.3.4.1.3.4

Multiply $2$ by $1$ .

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

Step 4.3.4.1.3.5

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

$0 + 2 e^{2 t}$

Step 4.3.4.1.4

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

$2 e^{2 t}$

Step 4.3.4.2

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

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

Step 4.3.5

Simplify.

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Step 4.3.5.1

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

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

Step 4.3.5.2

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

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

Step 4.3.6

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

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

Step 4.3.7

Simplify.

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Step 4.3.7.1

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

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

Step 4.3.7.2

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

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Step 4.3.7.2.1

Factor $2$ out of $- 2$ .

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

Step 4.3.7.2.2

Cancel the common factors.

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Step 4.3.7.2.2.1

Factor $2$ out of $2$ .

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

Step 4.3.7.2.2.2

Cancel the common factor.

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

Step 4.3.7.2.2.3

Rewrite the expression.

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

Step 4.3.7.2.2.4

Divide $- 1$ by $1$ .

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

Step 4.3.8

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

$\ln (| x |) + C_{1} = - (\ln (| u_{2} |) + C_{2})$

Step 4.3.9

Simplify.

$\ln (| x |) + C_{1} = - \ln (| u_{2} |) + C_{2}$

Step 4.3.10

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

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

Step 4.4

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

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