**Example 1:** Prove that x^2+x+1 is irreducible.

**Proof:** Since the discriminant is 1^2 - 4 \times 1 \times 1 < 0, it is irreducible.

**Problem 2:** Find \int \frac{dx}{x^2-a^2}, (a\neq 0).

**Hint:** Use \frac{1}{x^2-a^2}=\frac{1}{2a}\frac{1}{x-a}-\frac{1}{2a}\frac{1}{x+a}.

**Problem 3:** Find \int\frac{x^4-2x^2+4x+1}{x^3-x^2-x+1}dx.

**Hint:** Long division gives \frac{x^4-2x^2+4x+1}{x^3-x^2-x+1}dx = x + 1 + \frac{4x}{x^3-x^2-x+1}. Note that x^3-x^2-x+1 = (x-1)^2(x+1). The form of the partial fraction decomposition is \frac{4x}{(x-1)^2(x+1)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}\frac{C}{x+1}.

**Problem 4:** Find \int\frac{1}{x^2+a^2}dx.

**Hint:** Use x=au and dx=adu.

**Problem 5:** Find the partial fraction decomposition of \frac{t^6+1}{t^6+t^3}.

**Solution:** Long division gives \frac{t^6+1}{t^6+t^3} = 1+ \frac{-t^3+1}{t^6+t^3}. Note t^6+t^3=t^3(t^3+1)=t^3(t+1)(t^2-t+1). Note that \frac{-t^3+1}{t^6+t^3}=\frac{-(t^3+1)+2}{t^3(t^3+1)}=\frac{-1}{t^3}+\frac{2}{t^3(t^3+1)}=\frac{-1}{t^3}+\frac{2}{t^3}-\frac{2}{t^3+1}=\frac{1}{t^3}-\frac{2}{t^3+1}. Moreover the form of partial fraction decomposition of -\frac{2}{t^3+1} is \frac{A}{t+1}+\frac{Bx+C}{t^2-t+1}. Solve A=-\frac{2}{3}, B=\frac{2}{3} , C=-\frac{4}{3}. Therefore \frac{t^6+1}{t^6+t^3} = 1 + \frac{1}{t^3} - \frac{2}{3}\frac{1}{t+1} + \frac{2}{3}\frac{t-2}{t^2-t+1}.