Answer
$$\int\frac{y^4+y^2-1}{y^3+y}dy=\frac{y^2}{2}-\ln|y|+\frac{1}{2}\ln|y^2+1|+C$$
Work Step by Step
$$I=\int\frac{y^4+y^2-1}{y^3+y}dy$$
1) Perform long division:
We perform long division to rewrite the fraction as a polynomial plus a proper fraction, which is $$\frac{y^4+y^2-1}{y^3+y}=y+\frac{(-1)}{y^3+y}=y+\frac{-1}{y(y^2+1)}$$
2) Express the remainder fraction as a sum of partial fractions:
$$\frac{-1}{y(y^2+1)}=\frac{A}{y}+\frac{By+C}{y^2+1}$$ (since $y^2+1$ is irreducible)
Clear fractions:
$$A(y^2+1)+(By+C)y=-1$$ $$Ay^2+A+By^2+Cy=-1$$ $$(A+B)y^2+Cy+A=-1$$
Equating coefficients of corresponding powers of $y$, we get
$C=0$ and $A=-1$
$A+B=0$, so $B=1$
Therefore,
$$\frac{-1}{y(y^2+1)}=-\frac{1}{y}+\frac{y}{y^2+1}$$
2) Evaluate the integral:
$$I=\int ydy-\int\frac{1}{y}dy+\int\frac{y}{y^2+1}dy$$ $$I=\frac{y^2}{2}-\ln|y|+\frac{1}{2}\int\frac{1}{y^2+1}d(y^2+1)$$ $$I=\frac{y^2}{2}-\ln|y|+\frac{1}{2}\ln|y^2+1|+C$$
