Calculus Examples

$\frac{d y}{d x} = {(e^{x} + 8)}^{2}$

Step 1

Rewrite the equation.

$d y = {(e^{x} + 8)}^{2} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int {(e^{x} + 8)}^{2} d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int {(e^{x} + 8)}^{2} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Simplify.

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Step 2.3.1.1

Rewrite ${(e^{x} + 8)}^{2}$ as $(e^{x} + 8) (e^{x} + 8)$ .

$y + C_{1} = \int (e^{x} + 8) (e^{x} + 8) d x$

Step 2.3.1.2

Expand $(e^{x} + 8) (e^{x} + 8)$ using the FOIL Method.

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Step 2.3.1.2.1

Apply the distributive property.

$y + C_{1} = \int e^{x} (e^{x} + 8) + 8 (e^{x} + 8) d x$

Step 2.3.1.2.2

Apply the distributive property.

$y + C_{1} = \int e^{x} e^{x} + e^{x} \cdot 8 + 8 (e^{x} + 8) d x$

Step 2.3.1.2.3

Apply the distributive property.

$y + C_{1} = \int e^{x} e^{x} + e^{x} \cdot 8 + 8 e^{x} + 8 \cdot 8 d x$

Step 2.3.1.3

Simplify and combine like terms.

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Step 2.3.1.3.1

Simplify each term.

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Step 2.3.1.3.1.1

Multiply $e^{x}$ by $e^{x}$ by adding the exponents.

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Step 2.3.1.3.1.1.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y + C_{1} = \int e^{x + x} + e^{x} \cdot 8 + 8 e^{x} + 8 \cdot 8 d x$

Step 2.3.1.3.1.1.2

Add $x$ and $x$ .

$y + C_{1} = \int e^{2 x} + e^{x} \cdot 8 + 8 e^{x} + 8 \cdot 8 d x$

Step 2.3.1.3.1.2

Move $8$ to the left of $e^{x}$ .

$y + C_{1} = \int e^{2 x} + 8 \cdot e^{x} + 8 e^{x} + 8 \cdot 8 d x$

Step 2.3.1.3.1.3

Multiply $8$ by $8$ .

$y + C_{1} = \int e^{2 x} + 8 e^{x} + 8 e^{x} + 64 d x$

Step 2.3.1.3.2

Add $8 e^{x}$ and $8 e^{x}$ .

$y + C_{1} = \int e^{2 x} + 16 e^{x} + 64 d x$

Step 2.3.2

Split the single integral into multiple integrals.

$y + C_{1} = \int e^{2 x} d x + \int 16 e^{x} d x + \int 64 d x$

Step 2.3.3

Let $u = 2 x$ . Then $d u = 2 d x$ , so $\frac{1}{2} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.3.1

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

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Step 2.3.3.1.1

Differentiate $2 x$ .

$\frac{d}{d x} [2 x]$

Step 2.3.3.1.2

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

$2 \frac{d}{d x} [x]$

Step 2.3.3.1.3

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

$2 \cdot 1$

Step 2.3.3.1.4

Multiply $2$ by $1$ .

$2$

Step 2.3.3.2

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

$y + C_{1} = \int e^{u} \frac{1}{2} d u + \int 16 e^{x} d x + \int 64 d x$

Step 2.3.4

Combine $e^{u}$ and $\frac{1}{2}$ .

$y + C_{1} = \int \frac{e^{u}}{2} d u + \int 16 e^{x} d x + \int 64 d x$

Step 2.3.5

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

$y + C_{1} = \frac{1}{2} \int e^{u} d u + \int 16 e^{x} d x + \int 64 d x$

Step 2.3.6

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

$y + C_{1} = \frac{1}{2} (e^{u} + C_{2}) + \int 16 e^{x} d x + \int 64 d x$

Step 2.3.7

Since $16$ is constant with respect to $x$ , move $16$ out of the integral.

$y + C_{1} = \frac{1}{2} (e^{u} + C_{2}) + 16 \int e^{x} d x + \int 64 d x$

Step 2.3.8

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

$y + C_{1} = \frac{1}{2} (e^{u} + C_{2}) + 16 (e^{x} + C_{3}) + \int 64 d x$

Step 2.3.9

Apply the constant rule.

$y + C_{1} = \frac{1}{2} (e^{u} + C_{2}) + 16 (e^{x} + C_{3}) + 64 x + C_{4}$

Step 2.3.10

Simplify.

$y + C_{1} = \frac{1}{2} e^{u} + 16 e^{x} + 64 x + C_{5}$

Step 2.3.11

Replace all occurrences of $u$ with $2 x$ .

$y + C_{1} = \frac{1}{2} e^{2 x} + 16 e^{x} + 64 x + C_{5}$

Step 2.3.12

Reorder terms.

$y + C_{1} = \frac{1}{2} e^{2 x} + 64 x + 16 e^{x} + C_{5}$

Step 2.4

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

$y = \frac{1}{2} e^{2 x} + 64 x + 16 e^{x} + K$