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:)
 A: 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}
A: If you want to use partial fractions for a solution to something, then you first have to check the numerator and denominator's degrees (the degree of a polynomial is the highest power in it, eg the degree of $x^3+5x-4$ is $3$).

*

*If the degree of the numerator is $\ge$ the degree of the denominator, then before using partial fractions we must use long division or some other algebraic technique to transform the fraction we're dealing with into a fraction with the degree of the numerator that is less than the degree of the denominator, and a remainder term. For example,
$$\frac{x^2}{x^2-1}=\frac{x^2-1+1}{x^2-1}=1+\frac{1}{(x+1)(x-1)}$$
and now we're in a position to use partial fractions.


*If the degree of the denominator is $>$ than the degree of the numerator than we can immediately proceed with partial fractions.
Your integral fits into the first category, so you have a bit more work to do before you can use partial fractions.
