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

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

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

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

diverges

Solution steps

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

Find undefined (singularity) points:

x = 0

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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