What is the integral from 1 to 2 of sqrt(1+(1/x)^2) ?

The integral from 1 to 2 of sqrt(1+(1/x)^2) is -sqrt(2)+ln(sqrt(2)+1)+sqrt(5)-ln((sqrt(5)+1)/2)

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integral from 1 to 2 of sqrt(1+(1/x)^2)

\int_{1}^{2} 1 + (\frac{1}{x})^{2} d x

- 2 + ln (2 + 1) + 5 - ln (\frac{5 + 1}{2})

Decimal

1.22201 \dots

Solution steps

\int_{1}^{2} 1 + (\frac{1}{x})^{2} d x

Apply u-substitution

: \int_{1}^{\frac{1}{2}} - \frac{1 + u ^{2}}{u ^{2}} d u

= \int_{1}^{\frac{1}{2}} - \frac{1 + u ^{2}}{u ^{2}} d u

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

= - \int_{\frac{1}{2}}^{1} - \frac{1 + u ^{2}}{u ^{2}} d u

Take the constant out:

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

= - (- \int_{\frac{1}{2}}^{1} \frac{1 + u ^{2}}{u ^{2}} d u)

Apply Integration By Parts

: [- \frac{1 + u ^{2}}{u} - \int - \frac{1}{1 + u ^{2}} d u]_{\frac{1}{2}}^{1}

= - (- [- \frac{1 + u ^{2}}{u} - \int - \frac{1}{1 + u ^{2}} d u]_{\frac{1}{2}}^{1})

\int - \frac{1}{1 + u ^{2}} d u = - ln u^{2} + 1 + u

= - (- [- \frac{1 + u ^{2}}{u} - (- ln u^{2} + 1 + u)]_{\frac{1}{2}}^{1})

Simplify

= [- \frac{1 + u ^{2}}{u} + ln u^{2} + 1 + u]_{\frac{1}{2}}^{1}

Compute the boundaries:

- 2 + ln (2 + 1) + 5 - ln (\frac{5}{2} + \frac{1}{2})

= - 2 + ln (2 + 1) + 5 - ln (\frac{5}{2} + \frac{1}{2})

Simplify

= - 2 + ln (2 + 1) + 5 - ln (\frac{5 + 1}{2})

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What is the integral from 1 to 2 of sqrt(1+(1/x)^2) ?
The integral from 1 to 2 of sqrt(1+(1/x)^2) is -sqrt(2)+ln(sqrt(2)+1)+sqrt(5)-ln((sqrt(5)+1)/2)