What is the integral from-2 to 1 of 1/(x^4) ?

The integral from-2 to 1 of 1/(x^4) is diverges

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integral from-2 to 1 of 1/(x^4)

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

diverges

Solution steps

\int_{- 2}^{1} \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_{- 2}^{0} \frac{1}{x ^{4}} d x + \int_{0}^{1} \frac{1}{x ^{4}} d x

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{\partial}{\partial x} (2 y^{2} + 3 x)

\int \frac{x ^{10}}{x ^{2} + x - 2} d x

\int \frac{x ^{2}}{( x ^{3} - 2 ) ^{3}} d x

\frac{d}{d x} (\frac{8 x ^{2} + 4 x + 8}{x})

4 x^{3}, at x = 1

What is the integral from-2 to 1 of 1/(x^4) ?
The integral from-2 to 1 of 1/(x^4) is diverges