What is the integral from 0 to 4 of 1/(x^4-16) ?

The integral from 0 to 4 of 1/(x^4-16) is diverges

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integral from 0 to 4 of 1/(x^4-16)

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

diverges

Solution steps

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

Find undefined (singularity) points:

x = 2

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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\int_{0}^{l n (3)} 3 e^{3 x} d x

What is the integral from 0 to 4 of 1/(x^4-16) ?
The integral from 0 to 4 of 1/(x^4-16) is diverges