What is the integral from-3 to 4 of 8/(x^3) ?

The integral from-3 to 4 of 8/(x^3) is diverges

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integral from-3 to 4 of 8/(x^3)

\int_{- 3}^{4} \frac{8}{x ^{3}} d x

diverges

Solution steps

\int_{- 3}^{4} \frac{8}{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_{- 3}^{0} \frac{8}{x ^{3}} d x + \int_{0}^{4} \frac{8}{x ^{3}} d x

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d y}{d x} + \frac{1}{x} y = 5 x^{2}

\frac{d}{d x} (\frac{a + b x}{a - b x})

y^{^{'}} + 10 y = t + e^{- 5 t}

\frac{d}{d x} (x^{2} \cdot cos (x^{2}))

\int (e^{- x} + 4^{x}) d x

What is the integral from-3 to 4 of 8/(x^3) ?
The integral from-3 to 4 of 8/(x^3) is diverges