<s>[INST] What did I do wrong with this integral? PROBLEM
I have to compute this indefinite integral:
$$\int(2-x)^2\ln(4x)\,\,dx$$

MY ATTEMPT
So I did integration by parts:
$$\begin{matrix}
u=(2-x)^2 & dv=\ln(4x)\,\,dx\\
du=(2x-4)\,\,dx & v=\int\ln(4x)\\
\end{matrix}$$
$$\int\ln(4x)\,\,dx=\;\;?$$
$$\begin{matrix}
a=\ln(4x) & db=dx\\
da={1\over x}\,\,dx & b=x\\
\end{matrix}$$
$$\int\ln(4x)\,\,dx=x\ln(4x)-\int x\,\cdot\frac1x\,\,dx$$
$$\int\ln(4x)\,\,dx=x\ln(4x)-x$$
So because of that:
$$\int(2-x)^2\ln(4x)\,\,dx=(2-x)^2\Big(x\ln(4x)-x\Big)-\int(2x-4)\Big(x\ln(4x)-x\Big)\,\,dx$$
Now integrating by parts one more time:
$$\begin{matrix}
u=2x-4 & dv=\Big(x\ln(4x)-x\Big)\,\,dx\\
du=2\,\,dx & v=\int\Big(x\ln(4x)-x\Big)\,\,dx\\
\end{matrix}$$
$$\int\Big(x\ln(4x)-x\Big)\,\,dx=\int x\ln(4x)\,\,dx\;-\int x\,\,dx$$
$$\begin{matrix}
a=\ln(4x) & db=x\,\,dx\\
da={1\over x}\,\,dx & b=\frac{x^2}{2}\\
\end{matrix}$$
$$\int x\ln(4x)\,\,dx=\frac{x^2}{2}\ln(4x)\;-\int\frac{x^2}{2}\,\cdot\frac1x\,\,dx$$
$$\int x\ln(4x)\,\,dx=\frac{x^2}{2}\ln(4x)\;-\int\frac x2\,\,dx$$
$$\int x\ln(4x)\,\,dx=\frac{x^2}{2}\ln(4x)\;-\frac{x^2}{4}$$
$$\int\Big(x\ln(4x)-x\Big)\,\,dx=\frac{x^2}{2}\ln(4x)\;-\frac{x^2}{4}-\frac{x^2}{2}$$
Finally, after all this mess, plugging in in the equation of the original integral:
$$\int(2-x)^2\ln(4x)\,\,dx=(2-x)^2\Big(x\ln(4x)-x\Big)-\frac{x^2}{2}\ln(4x)\;+\frac{x^2}{4}+\frac{x^2}{2}+c$$
Now, I cleaned up this mess a little to give a nicer answer:
$$(2-x)^2\Big(x\ln(4x)-x\Big)-\frac{x^2}{2}\ln(4x)\;+\frac{x^2}{4}+\frac{x^2}{2}+c$$
$$x(x^2-4x+4)\Big(\ln(4x)-1\Big)-\frac{x^2}{2}\ln(4x)\;+\frac{x^2}{4}+\frac{x^2}{2}+c$$
$$(x^3-4x^2+4x)\ln(4x)-x^2+4x-4-\frac{x^2}{2}\ln(4x)\;+\frac34 x^2+c$$
$$(x^3-4x^2+4x)\ln(4x)-\frac{x^2}{2}\ln(4x)-x^2+4x-4\;+\frac34 x^2+c$$
$$\boxed{\Big(x^3-\frac{9}{2}x^2+4x\Big)\ln(4x)-\frac{1}{4}x^2+4x-4+c}$$

ANSWER FROM WOLFRAMALPHA
WolframAlpha gives me a different answer:
$$\frac{1}{9}x\Big(-x^2+3(x^2-6x+12)\ln(4x)+9x-36\Big)+c$$
Even after rearranging the WA's answer, I get similar answer but the coefficients and other details don't match. I won't bother rearranging the WolframAlpha's answer here, but trust me, after putting it into similar form to my answer, it doesn't work.

So what did I do wrong?
 [/INST] Up to here
$$\int(2-x)^2\ln(4x)\,\,dx=(2-x)^2\Big(x\ln(4x)-x\Big)-\int(2x-4)\Big(x\ln(4x)-x\Big)\,\,dx$$
it seems to be correct but this
$$\int(2-x)^2\ln(4x)\,\,dx=(2-x)^2\Big(x\ln(4x)-x\Big)-\frac{x^2}{2}\ln(4x)\;+\frac{x^2}{4}+\frac{x^2}{2}+c$$
seems not correct since you didn't considered the term
$$A=(2x-4)\left(\frac{x^2}{2}\ln(4x)\;-\frac{x^2}{4}-\frac{x^2}{2}\right)$$
and also the calculation of this part $\left(\frac{x^2}{2}\ln(4x)\;-\frac{x^2}{4}-\frac{x^2}{2}\right)$ seems uncorrect indeed the integral should be
$$\int(2x-4)\Big(x\ln(4x)-x\Big)\,\,dx=-\frac89x^3+3x^2+\frac23(x^3-3x^2)\ln (4x)$$
Let try to fix form here, since
$$\int\Big(x\ln(4x)-x\Big)\,\,dx=\frac{x^2}{2}\ln(4x)\;-\frac{x^2}{4}-\frac{x^2}{2}=\frac{x^2}{2}\ln(4x)\;-\frac{3x^2}{4}$$
we have that
$$\int(2x-4)\Big(x\ln(4x)-x\Big)\,\,dx=\\
=(2x-4)\left(\frac{x^2}{2}\ln(4x)\;-\frac{3x^2}{4}\right)-2\int\Big(\frac{x^2}{2}\ln(4x)\;-\frac{3x^2}{4}\Big)\,\,dx=\\
=(2x-4)\left(\frac{x^2}{2}\ln(4x)\;-\frac{3x^2}{4}\right)+\frac{x^3}{2}-\int x^2\ln(4x)\,\,dx=\\
=(x^3-2x^2)\ln(4x)\;-(x^3-3x^2)-\frac13x^3\log (4x)+\frac19x^3=\\
-\frac89x^3+3x^2+\frac23(x^3-3x^2)\ln (4x)$$
</s>