What is the integral from 0 to 1 of 5(e^{1/x})/(x^3) ?

The integral from 0 to 1 of 5(e^{1/x})/(x^3) is diverges

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integral from 0 to 1 of 5(e^{1/x})/(x^3)

\int_{0}^{1} 5 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

diverges

Solution steps

\int_{0}^{1} 5 \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

= 5 \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}

= 5 [\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}

= 5 [\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

5 [\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} : 5 [\frac{- e ^{\frac{1}{x}} + e ^{\frac{1}{x}} x}{x}]_{0}^{1}

= 5 [\frac{- e ^{\frac{1}{x}} + e ^{\frac{1}{x}} x}{x}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

(\frac{8 x + 5}{8 x - 5})^{\frac{1}{2}}

e^{r} + r^{e}

f (x) = \frac{2 x}{3 - tan ( x )}

\frac{\partial}{\partial x} (sin (8 x - 9 y))

\frac{x}{2} + 3

What is the integral from 0 to 1 of 5(e^{1/x})/(x^3) ?
The integral from 0 to 1 of 5(e^{1/x})/(x^3) is diverges