What is the integral from-2 to 3 of (37)/(x^4) ?

The integral from-2 to 3 of (37)/(x^4) is diverges

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integral from-2 to 3 of (37)/(x^4)

\int_{- 2}^{3} \frac{37}{x ^{4}} d x

diverges

Solution steps

\int_{- 2}^{3} \frac{37}{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{37}{x ^{4}} d x + \int_{0}^{3} \frac{37}{x ^{4}} d x

\int_{- 2}^{0} \frac{37}{x ^{4}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

x \to 0 lim (\frac{20 sin ( 3 x )}{x})

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

h \to 0 lim \frac{( 5 ( x + h ) + 4 ) - 5 x + 4}{h}

4 x + 3, [0, 1]

\int \frac{3 x ^{3} - 2 x ^{2} + 6 x}{x} d x

What is the integral from-2 to 3 of (37)/(x^4) ?
The integral from-2 to 3 of (37)/(x^4) is diverges