Task with convergence: $\int_{0}^{+\infty }\frac{e^{\arctan x}}{x^{2}}dx$ and $\int_{0}^{+\infty }\frac{e^{\arctan x}}{1+x^{2}}dx$ a) $$\int_{0}^{+\infty }\frac{e^{\arctan x}}{x^{2}}dx$$
b) $$\int_{0}^{+\infty }\frac{e^{\arctan x}}{1 + x^{2}}dx$$
I know that is same , but I need to first check convergence which I think that when I put infinity $$e^{\frac{\pi}{2}}$$ and then that get out and I have 
 $$\int_{0}^{+\infty }\frac{1}{x^{2}}dx$$ that convergent because $p>1$
And now I need to calculate it.
And there is problem, I don't know how.
 A: There is a problem with
$$
\int_0^{+\infty}\frac{1}{x^2}\,dx.
$$
The problem is at $x=0$. The integral
$$
\int_0^1\frac{1}{x^2}\,dx=\lim_{a\to 0^+}\Bigl[-\frac{1}{x}\Bigr]_a^1=\cdots
$$
diverges, and so
$$
\int_0^{+\infty}\frac{1}{x^2}\,dx
$$
also diverges.
A: HINT:
The first integral has a singularity at $x=0$.  Inasmuch as we have
$$\frac{e^{\arctan x}}{x^2}\ge \frac{1}{x^2}$$
and
$$\int_{\epsilon}^{\infty}\frac{1}{x^2}\,dx=\frac{1}{\epsilon}\to \infty$$
as $\epsilon \to 0^+$, the integral diverges.
The second integral is convergent since there is no singularity at the origin.  We have
$$\frac{e^{\arctan x}}{1+x^2}\le \frac{e^{\pi/2}}{1+x^2}$$
and 
$$\int_0^{\infty}\frac{1}{1+x^2}\,dx=\frac{\pi}{2}$$
A: Considering the general case of $$I=\int\frac{e^{\tan^{-1}( x)}}{a + x^{2}}dx$$ developing the integrand as a Taylor series at $x=0$ leads to $$\frac{e^{\tan^{-1}( x)}}{a + x^{2}}=\frac{1}{a}+\frac{x}{a}+\frac{(a-2) x^2}{2 a^2}-\frac{(a+6) x^3}{6
   a^2}+O\left(x^4\right)$$ which shows the problem if $a=0$ as already said in comments and answers.
Doing the same for infinitely large values of $x$ leads to $$\frac{e^{\tan^{-1}( x)}}{a + x^{2}}=\frac{e^{\pi /2}}{x^2}-\frac{e^{\pi /2}}{x^3}-\frac{e^{\pi /2} (2 a-1)}{2
   x^4}+O\left(\left(\frac{1}{x}\right)^5\right)$$
