What is the integral from 1 to 9 of (e^{1/x})/(x^2) ?

The integral from 1 to 9 of (e^{1/x})/(x^2) is e-\sqrt[9]{e}

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integral from 1 to 9 of (e^{1/x})/(x^2)

\int_{1}^{9} \frac{e ^{\frac{1}{x}}}{x ^{2}} d x

e - 9 e

Decimal

1.60076 \dots

Solution steps

\int_{1}^{9} \frac{e ^{\frac{1}{x}}}{x ^{2}} d x

Apply u-substitution

: \int_{1}^{\frac{1}{9}} - e^{u} d u

= \int_{1}^{\frac{1}{9}} - e^{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{\frac{1}{9}}^{1} - e^{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{\frac{1}{9}}^{1} e^{u} d u)

Use the common integral:

\int e^{u} d u = e^{u}

= - (- [e^{u}]_{\frac{1}{9}}^{1})

Simplify

= [e^{u}]_{\frac{1}{9}}^{1}

Compute the boundaries:

e - 9 e

= e - 9 e

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What is the integral from 1 to 9 of (e^{1/x})/(x^2) ?
The integral from 1 to 9 of (e^{1/x})/(x^2) is e-\sqrt[9]{e}