What is the integral from-1 to 8 of 1/(x^3) ?

The integral from-1 to 8 of 1/(x^3) is diverges

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integral from-1 to 8 of 1/(x^3)

\int_{- 1}^{8} \frac{1}{x ^{3}} d x

diverges

Solution steps

\int_{- 1}^{8} \frac{1}{x ^{3}} 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 ^{3}} d x + \int_{0}^{8} \frac{1}{x ^{3}} d x

\int_{- 1}^{0} \frac{1}{x ^{3}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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

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\int_{0}^{1} 2 + 3 x d x

\int_{0}^{1} (x^{3} - 2 x^{2} + x + 2) d x

\int_{0}^{2 π} 1 - ∣ cos (x) ∣ d x

What is the integral from-1 to 8 of 1/(x^3) ?
The integral from-1 to 8 of 1/(x^3) is diverges