Question

perform long division on the integrand, write the proper fraction as a sum of partial fractions
\\(\int\frac{3y^{4}+6y^{2}-1}{y^{3}+2y}dy\\)
perform long division on the given integrand and write the proper fraction as a sum of partial frac
\\(\frac{3y^{4}+6y^{2}-1}{y^{3}+2y}=\square\\)
(simplify your answer. use integers or fractions for any numbers in the expression )

Explanation:

Step1: Perform long - division

Divide $3y^{4}+6y^{2}-1$ by $y^{3}+2y$.
$3y^{4}+6y^{2}-1 = 3y(y^{3}+2y)-1$. So, $\frac{3y^{4}+6y^{2}-1}{y^{3}+2y}=3y-\frac{1}{y^{3}+2y}$.

Step2: Factor the denominator of the proper fraction

Factor $y^{3}+2y = y(y^{2}+2)$.

Step3: Write as partial - fractions

Let $\frac{1}{y(y^{2}+2)}=\frac{A}{y}+\frac{By + C}{y^{2}+2}$. Then $1=A(y^{2}+2)+(By + C)y=(A + B)y^{2}+Cy + 2A$.
Equating coefficients:
$A + B = 0$, $C = 0$, $2A=1$. Solving $2A = 1$ gives $A=\frac{1}{2}$. Since $A + B = 0$, then $B=-\frac{1}{2}$ and $C = 0$.
So, $\frac{1}{y(y^{2}+2)}=\frac{1}{2y}-\frac{y}{2(y^{2}+2)}$.
And $\frac{3y^{4}+6y^{2}-1}{y^{3}+2y}=3y-\frac{1}{2y}+\frac{y}{2(y^{2}+2)}$

Answer:

$3y-\frac{1}{2y}+\frac{y}{2(y^{2}+2)}$