What is the integral from-2 to 3 of (40)/(x^4) ?

The integral from-2 to 3 of (40)/(x^4) is diverges

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integral from-2 to 3 of (40)/(x^4)

\int_{- 2}^{3} \frac{40}{x ^{4}} d x

diverges

Solution steps

\int_{- 2}^{3} \frac{40}{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{40}{x ^{4}} d x + \int_{0}^{3} \frac{40}{x ^{4}} d x

\int_{- 2}^{0} \frac{40}{x ^{4}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

y = x + 3

\frac{1}{s} - \frac{2}{( s + 3 ) ^{2} - 9}

\int \frac{2 x ^{2} - 1}{x ^{3} - x} d x

\int \frac{1}{x ^{4} + y ^{2}} d y

\int 7 e^{- 3 x} d x

What is the integral from-2 to 3 of (40)/(x^4) ?
The integral from-2 to 3 of (40)/(x^4) is diverges