What is the integral from 0 to 1 of sqrt(1+(-1/x)^2) ?

The integral from 0 to 1 of sqrt(1+(-1/x)^2) is diverges

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integral from 0 to 1 of sqrt(1+(-1/x)^2)

\int_{0}^{1} 1 + (- \frac{1}{x})^{2} d x

diverges

Solution steps

\int_{0}^{1} 1 + (- \frac{1}{x})^{2} d x

Simplify

1 + (- \frac{1}{x})^{2} : \frac{x ^{2} + 1}{x ^{2}}

= \int_{0}^{1} \frac{x ^{2} + 1}{x ^{2}} d x

Simplify

\frac{x ^{2} + 1}{x ^{2}} : \frac{x ^{2} + 1}{x ^{2}}

= \int_{0}^{1} \frac{x ^{2} + 1}{x ^{2}} d x

x^{2} = (x),

assuming

x \geq 0

= \int_{0}^{1} \frac{x ^{2} + 1}{x} d x

Apply u-substitution

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

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

Factor

u^{2} - 1 : - (- u^{2} + 1)

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

Take the constant out:

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

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

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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What is the integral from 0 to 1 of sqrt(1+(-1/x)^2) ?
The integral from 0 to 1 of sqrt(1+(-1/x)^2) is diverges