Recitation 6

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}.

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