What is the integral from 0 to 12 of 1/(36-x^2) ?

The integral from 0 to 12 of 1/(36-x^2) is diverges

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integral from 0 to 12 of 1/(36-x^2)

\int_{0}^{12} \frac{1}{36 - x ^{2}} d x

diverges

Solution steps

\int_{0}^{12} \frac{1}{36 - x ^{2}} d x

Find undefined (singularity) points:

x = 6

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}^{6} \frac{1}{36 - x ^{2}} d x + \int_{6}^{12} \frac{1}{36 - x ^{2}} d x

\int_{0}^{6} \frac{1}{36 - x ^{2}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{\frac{π}{6}} t sin (2 t) d t

\frac{4}{s ^{2} ( s + 2 )} + \frac{3}{s + 2}

\frac{d}{d x} (arctan (\frac{x}{4 - 16 - x ^{2}}))

x \to 1 lim (\frac{2 x}{6 x - 10})

\frac{d}{d x} (60)

What is the integral from 0 to 12 of 1/(36-x^2) ?
The integral from 0 to 12 of 1/(36-x^2) is diverges