# 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 $\frac{t^6+1}{t^6+t^3} = 1 + \frac{1}{t^3} - \frac{2}{3}\frac{1}{t+1} + \frac{2}{3}\frac{t-2}{t^2-t+1}$.