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$$int_0^4 frac{2x}{(4-x^2)^2} ,dx$$

Simple question, but when I evaluate this with u-substitution, I get: $frac{-1}{3}$

However, the answer states it evaluates to: $+infty$ which is divergent.

To clarify; is this divergent because there is a vertical asymptote at $x = 2$, and since the interval's range is from $0 : 4$, there exists a point where the function equals $frac{x}{0}$?

It is not just that there is a vertical asymptote in the interval you are integrating over, it is also that that specific singularity "blows up too fast". If you split that into two improper integrals and do the math, your answer will also be $infty$, but for fun you should compare the following two integrals:
$$
int_{-1}^1 frac{1}{x^2},dx
$$
$$
int_{-1}^1 frac{1}{|x|^{1/2}},dx
$$

No. The function$$begin{array}{rccc}fcolon&(0,1]&longrightarrow&mathbb R\&x&mapsto&dfrac1{sqrt x}end{array}$$also as a vertical asymptote at $0$; however, the integral $int_0^1f(x),mathrm dx$ converges

It is not necessary that the integral will diverge if there are vertical asymptotes in the domain of integration. Consider, for example, $int_{-1}^1frac1xdx$. The function $frac1x$ has a vertical asymptote at $x=0$, yet the integral converges to $0$.

To see why this is happening, split the integral at the vertical asymptote.

$int_{-1}^1frac1xdx=int_{-1}^0frac1xdx+int_0^1frac1xdx=ln|x|Big|_{x=-1}^{x=0}+ln|x|Big|_{x=0}^{x=1}$

Now note that in the first term, $x$ approaches $0$ from the left side, or $xto0^-$; in the second term, $x$ approaches $0$ from the right side, that is $xto0^+$. But in both cases, $|x|to0^+$, so the value of the integral $I=ln|x|Big|_{x=-1}^{x=0}+ln|x|Big|_{x=0}^{x=1}=-ln|-1|+ln1+ln|x|Big|_{xto0^+}^{xto0^-}=0$, because $ln|0^-|-ln|0^+|=lim_{hto0}ln h-ln h=0 (infty-infty$ form$)$.

So while $int_0^1frac1xdx$ "blows up", $int_{-1}^1frac1xdx$ is $0$ because the "negative" area to the left of $0$ also "blows up" at the same rate and neutralizes the "positive" area to its right.

Now, coming to your question, one thing you can note right away is that the integrand $frac{2x}{(4-x^2)^2}$ is non-negative in $[0,4]$, so the area under the curve cannot be $-1/3$. When you observe this, you know something fishy is going on.

Again, split up the integral at the vertical asymptote:

$int_0^4frac{2x}{(4-x^2)^2}dx=int_0^2frac{2x}{(4-x^2)^2}dx+int_2^4frac{2x}{(4-x^2)^2}dx=frac1{4-x^2}Big|_0^2+frac1{4-x^2}Big|_2^4$

You can see that both the terms of the integral diverge as was the case with $frac1x$, but carefully note the sign of the $infty$ they diverge to. In the first term, $xto2^-implies4-x^2to0^+impliesfrac1{4-x^2}to+infty$. In the second term, $xto2^+implies4-x^2to0^-impliesfrac1{4-x^2}to-infty$. So,

$frac1{4-x^2}Big|_0^2+frac1{4-x^2}Big|_2^4=-frac13+frac1{4-x^2}Big|_{xto2^+}^{xto2^-}=-frac13+lim_{hto0}frac1{4-(2-h)^2}-frac1{4-(2+h)^2}to+infty (infty+infty)$ form.

The crux of the answer is that it is not necessary for an integral to diverge merely because of the presence a vertical asymptote in the domain of integration. Both the rate of growth of the integral about the asymptote and the sign of the values of it takes in that neighbourhood contribute to the convergence of the integral.