They do not really claim that "$I$ not in $p_i$ for every $i$ implies $I$ not in $\cup_i p_i$" (1) is logically equivalent to "For each $i,$ there exists $x_i \in I$ such that $x_i \notin p_k$ for all $k\neq i.$"(2). First of all, you haven't specified what $k$, so how could these two be equivalent?

What is trully happening is that in trying to prove (1), they use (2) in an induction argument. In particular, they assume that the statement is true whenever we have $n-1$ prime ideals. So suppose that $I \not\subset \cup_{i=1}^n p_i$. Then certainly for every $k= 1, ..., n$ we have that $I \not\subset \cup_{j \neq k} p_j$. Then apply the induction hypothesis to get (2).

Logic problem: Atiyah-Macdonald 1.11
Hi I have the following integral:
$$\int \frac{2x}{x^2+6x+3}\, dx$$

I made some changes like:
$$\int \dfrac{2x+6-6}{x^2+6x+3}\, dx$$

then I have: 
$$\int \dfrac{2x+6}{x^2+6x+3}\, dx -\int\dfrac{6}{x^2+6x+3}\, dx$$

and thus: $$\ln(x^2+6x+3)-\int\dfrac{6}{x^2+6x+3}\, dx$$

Ok, I have decomposed $$\frac{2x}{x^2+6x+3} $$  in: $$ \frac{3+\sqrt6}{\sqrt6(x+\sqrt 6+3)} + \frac{3-\sqrt6}{\sqrt6 (-x+\sqrt6-3)}$$

How can I integrate this expressions?
