What is the integral from-1 to 0 of 6(e^{1/x})/(x^3) ?

The integral from-1 to 0 of 6(e^{1/x})/(x^3) is -(12)/e

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integral from-1 to 0 of 6(e^{1/x})/(x^3)

\int_{- 1}^{0} 6 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

- \frac{12}{e}

Decimal

- 4.41455 \dots

Solution steps

\int_{- 1}^{0} 6 \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

= 6 \cdot \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}

= 6 [\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}

= 6 [\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

6 [\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} : 6 [\frac{- e ^{\frac{1}{x}} + e ^{\frac{1}{x}} x}{x}]_{- 1}^{0}

= 6 [\frac{- e ^{\frac{1}{x}} + e ^{\frac{1}{x}} x}{x}]_{- 1}^{0}

Compute the boundaries:

- \frac{2}{e}

= 6 (- \frac{2}{e})

Simplify

= - \frac{12}{e}

12 - x - x^{2}, x + 4

y 1 = x^{3} + 4 x^{2} - 6 x - 9, y 2 = 3 x^{2}

f (x) = e^{7 x^{2}}, at x = 2

\int y - y e^{z} d x

\int \frac{r}{( r ^{2} + x ^{2} ) ^{\frac{3}{2}}} d r

What is the integral from-1 to 0 of 6(e^{1/x})/(x^3) ?
The integral from-1 to 0 of 6(e^{1/x})/(x^3) is -(12)/e