What is the integral from-2 to 3 of (47)/(x^4) ?

The integral from-2 to 3 of (47)/(x^4) is diverges

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integral from-2 to 3 of (47)/(x^4)

\int_{- 2}^{3} \frac{47}{x ^{4}} d x

diverges

Solution steps

\int_{- 2}^{3} \frac{47}{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{47}{x ^{4}} d x + \int_{0}^{3} \frac{47}{x ^{4}} d x

\int_{- 2}^{0} \frac{47}{x ^{4}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{\partial}{\partial t} (ln (t^{2} + 1))

\frac{1}{6 x}

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

\int x sin (15 - x) d x

\int \frac{1}{cos ( 2 x )} d x

What is the integral from-2 to 3 of (47)/(x^4) ?
The integral from-2 to 3 of (47)/(x^4) is diverges