Calculus long division $\int\frac{y^4+3y^2-1}{y^3+3y}\ dy$

Question

I have a problem like this in my homework and want to see how to go by doing this problem. I understand the long division, but cannot get the partial fraction part. $$\int\frac{y^4+3y^2-1}{y^3+3y}\ dy$$

Answer 1

HINT:

$$\frac{y^4+3y^2-1}{y^3+3y}=\frac{y^2(y^2+3)}{y(y^2+3)}-\frac1{y(y^2+3)}$$

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

For $\displaystyle\frac y{y^2(y^2+3)},$ set $\displaystyle y^2=u$ and then in the numerator $\displaystyle3=(u+3)-u$

Answer 2

$$\frac{y^4+3y^2-1}{y^3+3y}=y-\frac{1}{y(y^2+3)}$$ and $$\frac{1}{y(y^2+3)}=\frac{A}{y}+\frac{By+C}{y^2+3}\ .$$ Multiplying out, $$1=A(y^2+3)+y(By+C)$$ and equating coefficients gives $$A+B=0\ ,\quad C=0\ ,\quad 3A=1\ .$$ So $$\int\frac{y^4+3y^2-1}{y^3+3y}\,dy =\int y\,dy-\frac{1}{3}\int\frac{dy}{y}+\frac{1}{3}\int\frac{y}{y^2+3}\,dy$$ and I think you can take it from here.