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

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

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

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

diverges

Solution steps

\int_{0}^{1} \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]_{0}^{1}

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

\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})]_{0}^{1}

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

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

Compute the boundaries:

diverges

= diverges

(0, 2.5), (0.833, 0)

f (x) = 2 x^{3} - 2 x, at x = - 1

\frac{\partial}{\partial y} (sin (x + 2 y))

\int_{0}^{π} 2 π x sin (x) d x

\frac{\partial}{\partial x} ((x y)^{5})

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