Calculus Examples

\int \frac{e^{\frac{1}{x}}}{x^{2}} d x

$\int \frac{e^{\frac{1}{x}}}{x^{2}} d x$

Step 1

Let

u = \frac{1}{x}

$u = \frac{1}{x}$ . Then

d u = - \frac{1}{x^{2}} d x

$d u = - \frac{1}{x^{2}} d x$ , so

- d u = \frac{1}{x^{2}} d x

$- d u = \frac{1}{x^{2}} d x$ . Rewrite using

u

$u$ and

d

$d$

u

$u$ .

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Step 1.1

Let

u = \frac{1}{x}

$u = \frac{1}{x}$ . Find

\frac{d u}{d x}

$\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate

\frac{1}{x}

$\frac{1}{x}$ .

\frac{d}{d x} [\frac{1}{x}]

$\frac{d}{d x} [\frac{1}{x}]$

Step 1.1.2

Rewrite

\frac{1}{x}

$\frac{1}{x}$ as

x^{- 1}

$x^{- 1}$ .

\frac{d}{d x} [x^{- 1}]

$\frac{d}{d x} [x^{- 1}]$

Step 1.1.3

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = - 1

$n = - 1$ .

- x^{- 2}

$- x^{- 2}$

Step 1.1.4

Rewrite the expression using the negative exponent rule

b^{- n} = \frac{1}{b^{n}}

$b^{- n} = \frac{1}{b^{n}}$ .

- \frac{1}{x^{2}}

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

- \frac{1}{x^{2}}

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

Step 1.2

Rewrite the problem using

u

$u$ and

d u

$d u$ .

\int - e^{u} d u

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

\int - e^{u} d u

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

Step 2

Since

- 1

$- 1$ is constant with respect to

u

$u$ , move

- 1

$- 1$ out of the integral.

- \int e^{u} d u

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

Step 3

The integral of

e^{u}

$e^{u}$ with respect to

u

$u$ is

e^{u}

$e^{u}$ .

- (e^{u} + C)

$- (e^{u} + C)$

Step 4

Simplify.

- e^{u} + C

$- e^{u} + C$

Step 5

Replace all occurrences of

u

$u$ with

\frac{1}{x}

$\frac{1}{x}$ .

- e^{\frac{1}{x}} + C

$- e^{\frac{1}{x}} + C$

\int_{}^{} \frac{e^{\frac{1}{x}}}{x^{2}} d x

$\int_{}^{} \frac{e^{\frac{1}{x}}}{x^{2}} d x$

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