What is the integral from 0 to 8 of 1/(16-x^2) ?

The integral from 0 to 8 of 1/(16-x^2) is diverges

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integral from 0 to 8 of 1/(16-x^2)

\int_{0}^{8} \frac{1}{16 - x ^{2}} d x

diverges

Solution steps

\int_{0}^{8} \frac{1}{16 - x ^{2}} d x

Find undefined (singularity) points:

x = 4

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{4} \frac{1}{16 - x ^{2}} d x + \int_{4}^{8} \frac{1}{16 - x ^{2}} d x

\int_{0}^{4} \frac{1}{16 - x ^{2}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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What is the integral from 0 to 8 of 1/(16-x^2) ?
The integral from 0 to 8 of 1/(16-x^2) is diverges