Integrate: integral (1)/(x2sqrt(x2-9)) dx

The integral contains $\sqrt{x^2 - a^2}$ where $a=3$. Let $x = 3 \sec \theta$.
Then $dx = 3 \sec \theta \tan \theta \, d\theta$.
And $\sqrt{x^2 - 9} = \sqrt{(3 \sec \theta)^2 - 9} = \sqrt{9 \sec^2 \theta - 9} = \sqrt{9(\sec^2 \theta - 1)} = \sqrt{9 \tan^2 \theta} = 3 \tan \theta$.

From $x = 3 \sec \theta$, we have $\sec \theta = \frac{x}{3}$. This means $\cos \theta = \frac{3}{x}$.
Consider a right triangle where the hypotenuse is $x$ and the adjacent side is 3. The opposite side is $\sqrt{x^2 - 3^2} = \sqrt{x^2 - 9}$.
Therefore, $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{x^2 - 9}}{x}$.
Substitute this back into the result from Step 3.

The final answer is $\boxed{\frac{\sqrt{x^2 - 9}}{9x} + C}$.

10. a) For any three sets A, B and C prove that: $(A \cup B) - C = (A - C) \cup (B - C)$.

To prove this set equality, we will show that each side is a subset of the other.

Let $x \in (A \cup B) - C$.
By the definition of set difference, this means $x \in (A \cup B)$ and $x \notin C$.
By the definition of union, $x \in A$ or $x \in B$.
So, we have $(x \in A \text{ or } x \in B)$ and $x \notin C$.
Using the distributive property of "and" over "or", this is equivalent to:
$(x \in A \text{ and } x \notin C)$ or $(x \in B \text{ and } x \notin C)$.
By the definition of set difference, this means $(x \in A - C)$ or $(x \in B - C)$.
By the definition of union, this means $x \in (A - C) \cup (B - C)$.
Therefore, $(A \cup B) - C \subseteq (A - C) \cup (B - C)$.

Let $x \in (A - C) \cup (B - C)$.
By the definition of union, this means $x \in (A - C)$ or $x \in (B - C)$.
By the definition of set difference, this means $(x \in A \text{ and } x \notin C)$ or $(x \in B \text{ and } x \notin C)$.
Using the distributive property of "and" over "or", this is equivalent to:
$(x \in A \text{ or } x \in B)$ and $x \notin C$.
By the definition of union, this means $x \in (A \cup B)$.
So, we have $x \in (A \cup B)$ and $x \notin C$.
By the definition of set difference, this means $x \in (A \cup B) - C$.
Therefore, $(A - C) \cup (B - C) \subseteq (A \cup B) - C$.

Since we have shown that $(A \cup B) - C \subseteq (A - C) \cup (B - C)$ and $(A - C) \cup (B - C) \subseteq (A \cup B) - C$, the two sets are equal.
Thus, $\boxed{(A \cup B) - C = (A - C) \cup (B - C)}$ is proven.