Evaluate $\int_{1}^{2}\frac{x}{(x^2-3)^2}\mathrm dx$ 
Evaluate $$\int_{1}^{2}\frac{x}{(x^2-3)^2}\,\mathrm dx$$ 

So I tried solving this problem and arrived with $-{\dfrac{3}{4}}$. However, after trying to check it with my calculator it says math error? 
 A: $\sqrt{3}$ is a zero for the denominator in the interval $[1,2]$. Furthermore, it is a degree-two zero since the denominator is squared. 
So near $x=\sqrt{3}$, the function behaves like $\frac{1}{x^2}$ behaves near $0$. And any integral of $\frac{1}{x^2}$ over an interval that includes either side of $x=0$ is divergent.
So that is an informal argument that let's you see the integral is infinite without any calculation.

More formally, 
$$\begin{align}
\int_1^{\sqrt{3}}\frac{x}{(x^2-3)^2}\,dx
&=\lim_{t\to\sqrt3\,^-}\int_1^{t}\frac{x}{(x^2-3)^2}\,dx\\
&=\lim_{t\to\sqrt3\,^-}\left[-{\frac12}\frac1{x^2-3}\right]_1^t\\
&=\lim_{t\to\sqrt3\,^-}\left(-{\frac12}\frac1{t^2-3}-\frac14\right)\\
&=+\infty
\end{align}$$
So even without looking at the other side, the integral is divergent.
A: Evaluating indefinte integral is fine
$$I=\int\frac{x}{(x^2-3)^2}\,\mathrm dx=\frac12\int\frac{2x}{(x^2-3)^2}\,\mathrm dx$$
$$x^2-3=t\iff2x\,\mathrm dx=\,\mathrm dt$$
$$I=\frac12\int\frac{1}{t^2}\,\mathrm dt=-\frac{1}{2t}+C=-\frac{1}{2(x^2-3)}+C$$

But, Integrand has Non-integrable singularity of at $x=\sqrt3$

Indefinite integral $\displaystyle\int\frac{x}{(x^2-3)^2}\,\mathrm dx$ does not converge
