What is the integral from 0 to 1 of 7(e^{1/x})/(x^3) ?

The integral from 0 to 1 of 7(e^{1/x})/(x^3) is diverges

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integral from 0 to 1 of 7(e^{1/x})/(x^3)

\int_{0}^{1} 7 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

diverges

Solution steps

\int_{0}^{1} 7 \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

= 7 \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}

= 7 [\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}

= 7 [\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

7 [\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} : 7 [\frac{- e ^{\frac{1}{x}} + e ^{\frac{1}{x}} x}{x}]_{0}^{1}

= 7 [\frac{- e ^{\frac{1}{x}} + e ^{\frac{1}{x}} x}{x}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

\int \frac{1}{1 - 2 cos ( x )} d x

sin (x) + \frac{1}{9} cot (x)

\int \frac{2 x ^{5} + x}{x} d x

\int_{- 1}^{0} x d x

\frac{d y}{d x} = 7 y e^{x + 8}

What is the integral from 0 to 1 of 7(e^{1/x})/(x^3) ?
The integral from 0 to 1 of 7(e^{1/x})/(x^3) is diverges