What is the integral from-1 to 2 of (2x-7)/(x^3) ?

The integral from-1 to 2 of (2x-7)/(x^3) is diverges

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integral from-1 to 2 of (2x-7)/(x^3)

\int_{- 1}^{2} \frac{2 x - 7}{x ^{3}} d x

diverges

Solution steps

\int_{- 1}^{2} \frac{2 x - 7}{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{2 x - 7}{x ^{3}} d x + \int_{0}^{2} \frac{2 x - 7}{x ^{3}} d x

\int_{- 1}^{0} \frac{2 x - 7}{x ^{3}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{1}{1 - x}, 3

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

\frac{d}{d x} (\frac{1}{1 + 2 x})

\frac{d}{d x} (9 + 5 x)

\frac{d}{d x} (2 x^{- 7 3 x})

What is the integral from-1 to 2 of (2x-7)/(x^3) ?
The integral from-1 to 2 of (2x-7)/(x^3) is diverges