Calculus Examples

$\int {(e^{t} + 1)}^{2} d t$

Step 1

Simplify.

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Step 1.1

Rewrite ${(e^{t} + 1)}^{2}$ as $(e^{t} + 1) (e^{t} + 1)$ .

$\int (e^{t} + 1) (e^{t} + 1) d t$

Step 1.2

Expand $(e^{t} + 1) (e^{t} + 1)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

$\int e^{t} (e^{t} + 1) + 1 (e^{t} + 1) d t$

Step 1.2.2

Apply the distributive property.

$\int e^{t} e^{t} + e^{t} \cdot 1 + 1 (e^{t} + 1) d t$

Step 1.2.3

Apply the distributive property.

$\int e^{t} e^{t} + e^{t} \cdot 1 + 1 e^{t} + 1 \cdot 1 d t$

Step 1.3

Simplify and combine like terms.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

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

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Step 1.3.1.1.1

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

$\int e^{t + t} + e^{t} \cdot 1 + 1 e^{t} + 1 \cdot 1 d t$

Step 1.3.1.1.2

Add $t$ and $t$ .

$\int e^{2 t} + e^{t} \cdot 1 + 1 e^{t} + 1 \cdot 1 d t$

Step 1.3.1.2

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

$\int e^{2 t} + e^{t} + 1 e^{t} + 1 \cdot 1 d t$

Step 1.3.1.3

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

$\int e^{2 t} + e^{t} + e^{t} + 1 \cdot 1 d t$

Step 1.3.1.4

Multiply $1$ by $1$ .

$\int e^{2 t} + e^{t} + e^{t} + 1 d t$

Step 1.3.2

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

$\int e^{2 t} + 2 e^{t} + 1 d t$

Step 2

Split the single integral into multiple integrals.

$\int e^{2 t} d t + \int 2 e^{t} d t + \int d t$

Step 3

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

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Step 3.1

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

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Step 3.1.1

Differentiate $2 t$ .

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

Step 3.1.2

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

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

Step 3.1.3

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

$2 \cdot 1$

Step 3.1.4

Multiply $2$ by $1$ .

$2$

Step 3.2

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

$\int e^{u} \frac{1}{2} d u + \int 2 e^{t} d t + \int d t$

Step 4

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

$\int \frac{e^{u}}{2} d u + \int 2 e^{t} d t + \int d t$

Step 5

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

$\frac{1}{2} \int e^{u} d u + \int 2 e^{t} d t + \int d t$

Step 6

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

$\frac{1}{2} (e^{u} + C) + \int 2 e^{t} d t + \int d t$

Step 7

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

$\frac{1}{2} (e^{u} + C) + 2 \int e^{t} d t + \int d t$

Step 8

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

$\frac{1}{2} (e^{u} + C) + 2 (e^{t} + C) + \int d t$

Step 9

Apply the constant rule.

$\frac{1}{2} (e^{u} + C) + 2 (e^{t} + C) + t + C$

Step 10

Simplify.

$\frac{1}{2} e^{u} + 2 e^{t} + t + C$

Step 11

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

$\frac{1}{2} e^{2 t} + 2 e^{t} + t + C$