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

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

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

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

\frac{e ^{9} - e}{9}

Decimal

900.04062 \dots

Solution steps

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

Apply u-substitution

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

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

e^{9} - e

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

Simplify

= \frac{e ^{9} - e}{9}

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