What is the integral from 0 to 2 of (sqrt(4+x^2))/x ?

The integral from 0 to 2 of (sqrt(4+x^2))/x is diverges

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integral from 0 to 2 of (sqrt(4+x^2))/x

\int_{0}^{2} \frac{4 + x ^{2}}{x} d x

diverges

Solution steps

\int_{0}^{2} \frac{4 + x ^{2}}{x} d x

Apply u-substitution

: \int_{2}^{22} \frac{u ^{2}}{u ^{2} - 4} d u

= \int_{2}^{22} \frac{u ^{2}}{u ^{2} - 4} d u

Factor

u^{2} - 4 : - (- u^{2} + 4)

= \int_{2}^{22} \frac{u ^{2}}{- ( - u ^{2} + 4 )} d u

Take the constant out:

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

= - \int_{2}^{22} \frac{u ^{2}}{- u ^{2} + 4} d u

Apply Integral Substitution

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

Take the constant out:

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

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

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

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

Apply the Sum Rule:

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

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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What is the integral from 0 to 2 of (sqrt(4+x^2))/x ?
The integral from 0 to 2 of (sqrt(4+x^2))/x is diverges