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

The integral from-1 to 0 of (e^{1/x})/(x^3) is -2/e

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

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

- \frac{2}{e}

Decimal

- 0.73575 \dots

Solution steps

\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}

\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}

= [\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

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

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

Compute the boundaries:

- \frac{2}{e}

= - \frac{2}{e}

\int_{1}^{2} (x^{2} + \frac{1}{x ^{4}}) d x

\int_{0}^{\frac{π}{4}} cos (x) - sin (x) d x

\int_{0}^{\frac{π}{2}} cos (x) sin (x) d x

\int_{0}^{π} (1 - sin^{3} (x)) d x

\int_{\frac{π}{2}}^{π} sin (x) cos (x) d x

What is the integral from-1 to 0 of (e^{1/x})/(x^3) ?
The integral from-1 to 0 of (e^{1/x})/(x^3) is -2/e