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