What is the integral from-1 to 0 of 9(e^{1/x})/(x^3) ?

The integral from-1 to 0 of 9(e^{1/x})/(x^3) is -(18)/e

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integral from-1 to 0 of 9(e^{1/x})/(x^3)

\int_{- 1}^{0} 9 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

- \frac{18}{e}

Decimal Notation

- 6.62182 \dots

Solution steps

\int_{- 1}^{0} 9 \cdot \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int_{- 1}^{0} \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

Apply Integration By Parts

: [\frac{e ^{\frac{1}{x}}}{x ^{2}} - \int - \frac{e ^{\frac{1}{x}} ( 3 x + 1 )}{x ^{4}} d x]_{- 1}^{0}

= 9 [\frac{e ^{\frac{1}{x}}}{x ^{2}} - \int - \frac{e ^{\frac{1}{x}} ( 3 x + 1 )}{x ^{4}} d x]_{- 1}^{0}

\int - \frac{e ^{\frac{1}{x}} ( 3 x + 1 )}{x ^{4}} d x = \frac{3 e ^{\frac{1}{x}}}{x} - e^{\frac{1}{x}} + \frac{e ^{\frac{1}{x}}}{x ^{2}} - \frac{2 e ^{\frac{1}{x}}}{x}

= 9 [\frac{e ^{\frac{1}{x}}}{x ^{2}} - (\frac{3 e ^{\frac{1}{x}}}{x} - e^{\frac{1}{x}} + \frac{e ^{\frac{1}{x}}}{x ^{2}} - \frac{2 e ^{\frac{1}{x}}}{x})]_{- 1}^{0}

Simplify

9 [\frac{e ^{\frac{1}{x}}}{x ^{2}} - (\frac{3 e ^{\frac{1}{x}}}{x} - e^{\frac{1}{x}} + \frac{e ^{\frac{1}{x}}}{x ^{2}} - \frac{2 e ^{\frac{1}{x}}}{x})]_{- 1}^{0} : 9 [\frac{- e ^{\frac{1}{x}} + e ^{\frac{1}{x}} x}{x}]_{- 1}^{0}

= 9 [\frac{- e ^{\frac{1}{x}} + e ^{\frac{1}{x}} x}{x}]_{- 1}^{0}

Compute the boundaries:

- \frac{2}{e}

= 9 (- \frac{2}{e})

Simplify

= - \frac{18}{e}

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What is the integral from-1 to 0 of 9(e^{1/x})/(x^3) ?
The integral from-1 to 0 of 9(e^{1/x})/(x^3) is -(18)/e