What is the integral from-2 to 3 of (33)/(x^4) ?

The integral from-2 to 3 of (33)/(x^4) is diverges

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integral from-2 to 3 of (33)/(x^4)

\int_{- 2}^{3} \frac{33}{x ^{4}} d x

diverges

Solution steps

\int_{- 2}^{3} \frac{33}{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{33}{x ^{4}} d x + \int_{0}^{3} \frac{33}{x ^{4}} d x

\int_{- 2}^{0} \frac{33}{x ^{4}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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

x \to - 7 lim (\frac{x ^{2} - 5}{7 - x})

\int \frac{cos ( x )}{sin ( x ) + cos ( x )} d x

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

\frac{( 3 y ^{2} - t ^{2} )}{y ^{5}} (\frac{d y}{d t}) + \frac{t}{2 y ^{4}} = 0

What is the integral from-2 to 3 of (33)/(x^4) ?
The integral from-2 to 3 of (33)/(x^4) is diverges