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

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

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

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

diverges

Solution steps

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

= 9 [\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}

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

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

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

Compute the boundaries:

diverges

= diverges

\frac{\partial}{\partial y} (8 e^{yz} cos (x yz))

(4.8) (2.12)

\frac{d}{d x} (x e^{2 x})

\int sin^{2} (\frac{π}{4} + \frac{t}{2}) d t

x \to - 7 lim (x^{2} - 49)

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