Calculus Examples

$\int {(1 + \frac{1}{t})}^{3} (\frac{1}{t^{2}}) d t$

Step 1

Combine ${(1 + \frac{1}{t})}^{3}$ and $\frac{1}{t^{2}}$ .

$\int \frac{{(1 + \frac{1}{t})}^{3}}{t^{2}} d t$

Step 2

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

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Step 2.1

Let $u = 1 + \frac{1}{t}$ . Find $\frac{d u}{d t}$ .

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Step 2.1.1

Differentiate $1 + \frac{1}{t}$ .

$\frac{d}{d t} [1 + \frac{1}{t}]$

Step 2.1.2

Differentiate.

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Step 2.1.2.1

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

$\frac{d}{d t} [1] + \frac{d}{d t} [\frac{1}{t}]$

Step 2.1.2.2

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

$0 + \frac{d}{d t} [\frac{1}{t}]$

Step 2.1.3

Evaluate $\frac{d}{d t} [\frac{1}{t}]$ .

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Step 2.1.3.1

Rewrite $\frac{1}{t}$ as $t^{- 1}$ .

$0 + \frac{d}{d t} [t^{- 1}]$

Step 2.1.3.2

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

$0 - t^{- 2}$

Step 2.1.4

Simplify.

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Step 2.1.4.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$0 - \frac{1}{t^{2}}$

Step 2.1.4.2

Subtract $\frac{1}{t^{2}}$ from $0$ .

$- \frac{1}{t^{2}}$

Step 2.2

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

$\int - u^{3} d u$

Step 3

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

$- \int u^{3} d u$

Step 4

By the Power Rule, the integral of $u^{3}$ with respect to $u$ is $\frac{1}{4} u^{4}$ .

$- (\frac{1}{4} u^{4} + C)$

Step 5

Rewrite $- (\frac{1}{4} u^{4} + C)$ as $- \frac{1}{4} u^{4} + C$ .

$- \frac{1}{4} u^{4} + C$

Step 6

Replace all occurrences of $u$ with $1 + \frac{1}{t}$ .

$- \frac{1}{4} {(1 + \frac{1}{t})}^{4} + C$