Question on Convergence of Improper integrals Question is to check which of the following improper integrals are convergent?
$$\int _1^{\infty} \frac{dx}{\sqrt{x^2+2x+2}}$$
$$\int _0^5 \frac{dx}{x^2-5x+6}$$
$$\int _0^5\frac{dx}{\sqrt[3]{7x+2x^4}}$$
I was having a stupid mindset that :
"$\textbf{Improper integrals should have at least one limit infinite}$" 
and so i came to conclusion that given two integrals with finite limits are convergent and thus question is nonsense.
But sooner i realized that for two  integrals $$\int _0^5 \frac{dx}{x^2-5x+6}$$
$$\int _0^5\frac{dx}{\sqrt[3]{7x+2x^4}}$$
denominators have zeros in given interval $[0,5]$
So, then I decided that these integrals are actually improper :)
Now, I tried seeing $$\int _1^{\infty} \frac{dx}{\sqrt{x^2+2x+2}}$$ as limit of $$\int _1^{b} \frac{dx}{\sqrt{x^2+2x+2}}$$ with $b\rightarrow \infty$
$$\int _1^{b} \frac{dx}{\sqrt{x^2+2x+2}}=2\sqrt{x^2+2x+2}.\frac{1}{2x+2}=\sqrt{x^2+2x+2}.\frac{1}{x+1}=\sqrt{\frac{(x+1)^2+1}{(x+1)^2}}=\sqrt{1+\frac{1}{(x+1)^2}}$$
sorry for forgetting limits
which would imply that $$\int _1^{b} \frac{dx}{\sqrt{x^2+2x+2}}=\sqrt{1+\frac{1}{(b+1)^2}}-\sqrt{1+\frac{1}{(1+1)^2}}$$
and when $b\rightarrow \infty$ as $b$ is in denominator... this limit exists and  so,
$$\int _1^{\infty} \frac{dx}{\sqrt{x^2+2x+2}}=1-\sqrt{\frac{5}{4}}$$
As $x^2+2x+2$  is positive in $[1,\infty]$, the integral should be positive, I might have missed some signs in between but i am sure that this is convergent.. (I would be thankful if some one can help me to see where the problem is with signs)
Coming to second problem, $$\frac{1}{x^2-5x-6}=\frac{1}{x-3}-\frac{1}{x-2}$$
$$\int _0^5 \frac{dx}{x^2-5x+6}=\int _0^5 \big(\frac{1}{x-3}-\frac{1}{x-2}\big)$$
$$\int _0^5 \big(\frac{1}{x-3}-\frac{1}{x-2}\big)=\int _0^2 \big(\frac{1}{x-3}-\frac{1}{x-2}\big)+\int _2^3 \big(\frac{1}{x-3}-\frac{1}{x-2}\big)+\int _3^5 \big(\frac{1}{x-3}-\frac{1}{x-2}\big)$$
$$=\log (2-3)- \log(2-2) \dots$$ but then $\log 0$ is not defined 
SO, I would like to confirm that this integral is divergent.
Coming to third integral $$\int _0^5\frac{dx}{\sqrt[3]{7x+2x^4}}$$
I would like to see this as limit of $$\int _a^5\frac{dx}{\sqrt[3]{7x+2x^4}}$$
and $$\int _a^5\frac{dx}{\sqrt[3]{7x+2x^4}}=\frac{3}{2}(7x+2x^4)^{\frac{2}{3}}.\frac{1}{8x^3+7}$$
with similar calculations as i have done for first integral, I have seen that this integral also converges (I do not want to write that and kill readers patience :D)
I am sure if my idea for first integral is correct then third integral should also converge.
So, I would say that first and third integrals converge where as second integral do not converge.
I would be thankful if someone can verify my solution and please let me know if there are any gaps.
Thank You.
P.S : I would like to remind for users (who are as dumb as i am) that 
$\textbf{Improper integrals need not necessarily have infinite end points}$

Hint: For the issue concerning convergence it sufficient and necessary to understand only the two following types of integrals for $\alpha \in \mathbb{R}$:
$$ \int_{0}^1 x^\alpha d x , \qquad  \int_{1}^\infty x^\alpha d x.$$
E.g.
$$ \int_{1}^\infty ( (x+1)^2 + 1)^{-1/2} d x = \int_{0}^\infty ( x^2 + 1)^{-1/2} d x \geq C * \int_{1}^\infty x^{-1} dx  = \infty$$
