Integrate $\int\frac{6x^4+4x^2+8x+4}{(1-2x)^3}dx$ Integrate $$\int\frac{6x^4+4x^2+8x+4}{(1-2x)^3}dx$$
I tried $2\cdot\int\frac{3x^4+2x^2+4x+2}{(1-2x)^3}dx\:\:$ now after partial fractions  i have:
$$\begin{align}
&\>\>\>\>\>\frac{3x^4+2x^2+4x+2}{(1-2x)^3}\\&=\frac{A}{(1-2x)}+\frac{B}{(1-2x)^2}+\frac{C}{(1-2x)^3}\\[1ex]
&=A(1-2x)^2(1-2x)^3+B(1-2x)(1-2x)^3+C(1-2x)(1-2x)^2\\[1ex]
&=(1-2x)^3\cdot[A(1-2x)^2+B(1-2x)+C]\\[1ex]
&=(1-2x)^3\cdot[A-4Ax+4Ax^2+B-2Bx+C]
\end{align}$$
Need a bit help to find out $A,B,C$ and evaluation of problem.
Thank you in advance:)

Substitute  $t=1-2x$. Then $x=\frac{1-t}2$
\begin{align}
& \int \frac{6x^4+4x^2+8x+4}{(1-2x)^3}dx\\
 =& \int \frac{6(\frac{1-t}2)^4+ 4(\frac{1-t}2)^2+ 8(\frac{1-t}2)+4}{t^3}(-\frac{dt}2)\\
=& -\frac1{16} \int \frac{3t^4 -12t^3+26t^2-60t+75}{t^3}dt
\end{align}
