What is the integral from-2 to 3 of (43)/(x^4) ?

The integral from-2 to 3 of (43)/(x^4) is diverges

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integral from-2 to 3 of (43)/(x^4)

\int_{- 2}^{3} \frac{43}{x ^{4}} d x

diverges

Solution steps

\int_{- 2}^{3} \frac{43}{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{43}{x ^{4}} d x + \int_{0}^{3} \frac{43}{x ^{4}} d x

\int_{- 2}^{0} \frac{43}{x ^{4}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

- 5 x^{3}

y^{^{''}} - 4 y^{^{'}} - 12 y = sin (2 t) + t e^{4 t}

\int - \frac{t}{t ^{2} + 1} d t

\int \frac{2}{1 - 4 x ^{2}} d x

\int_{- \frac{π}{4}}^{\frac{π}{4}} cos (2 x) d x

What is the integral from-2 to 3 of (43)/(x^4) ?
The integral from-2 to 3 of (43)/(x^4) is diverges