Answer
Divergent
Work Step by Step
Let \[I=\int_{-2}^{3}\:\frac{1}{x^4}\,dx\]
Since 0 is point of infinite discontinuity of integrand $\frac{1}{x^4}$ and $-2<0<3$
\[I=\int_{-2}^{0}\:\frac{1}{x^4}\,dx+\int_{0}^{3}\:\frac{1}{x^4}\,dx\;\;\;\ldots (1)\]
Let \[I_1=\int\frac{1}{x^4}dx=\int x^{-4}dx\]
\[I_1=\frac{-1}{3x^3}\;\;\;\ldots (2)\]
Let \[I_2=\int_{-2}^{0}\:\frac{1}{x^4}\,dx\]
\[I_2=\lim_{t_1\rightarrow 0^-}\int_{-2}^{t}\:\frac{1}{x^4}\,dx\;\;\;\ldots (3)\]
Using (2) in (3)
\[I_2=\lim_{t_1\rightarrow 0^-}\left[\frac{-1}{3x^3}\right]_{-2}^t\]
\[I_2=\lim_{t_1\rightarrow 0^-}\left[\frac{-1}{3t^3}+\frac{1}{3(-8)}\right]\]
$\;\;\;\;\;\;\;\;\;\;\;\Rightarrow $ does not exist
Since limit on R.H.S. of (3) does not exist so $I_2$ is divergent
From (1)
Consequently, $I$ is divergent.
