Integration by parts - is there an easier way to to set u and dv?

Question

The integral in question is $$\int(2-x)^2\ln{(4x)}dx$$

I've set $u=\ln{(4x)}$ with $du=\frac{4}{x}dx$ and $dv=(2-x)^2dx$ with $v=-\frac{1}{3}(2-x)^3$

Using integration by parts, I end up with:

$$-\frac{1}{3}(2-x)^3\ln{(4x)}-\int\frac{-4}{3x}(2-x)^3dx$$

$$=-\frac{1}{3}(2-x)^3\ln{(4x)}+\frac{32}{3}\ln{(x)}-16x+4x^2-\frac{4}{9}x^3+c$$

I don't know if I'm even correct at this point (don't have any answers), but I'm just wondering if it would be easier to set $u$ and $dv$ to something else. I've tried swapping the original $u$ and $dv$ but that seems like it would be more difficult.

Keep in mind, this is an integration by parts question, so using another method isn't really what I'm looking for.

EDIT: Thank you all, I now see my mistake.

Answer 1

The mistake is here: $$\left(\ln(4x)\right)'=\frac{1}{4x}\cdot (4x)'=\frac{1}{4x}\cdot 4=\frac1x.$$ So by your method: $$\int (2-x)^2\ln{(4x)}dx=\int \underbrace{\ln{(4x)}}_{u}\cdot\underbrace{(2-x)^2dx}_{dv}=\\ =\underbrace{\ln(4x)}_{u}\cdot \underbrace{\frac{-(2-x)^3}{3}}_{v}-\int \underbrace{\frac{-(2-x)^3}{3}}_{v}\cdot \underbrace{\frac1xdx}_{du}=\\ -\frac{(2-x)^3}{3}\cdot \ln(4x)+\frac{8}{3}\ln{(x)}-4x+x^2-\frac{1}{9}x^3+C.$$ It could be simplified as follows: $$I=\require{cancel}-\frac{8}{3}\cdot\ln(4x)+\left(4x-2x^2+\frac{x^3}{3}\right)\cdot \ln(4x)+\frac{8}{3}\ln{(x)}-4x+x^2-\frac{1}{9}x^3+C=\\ \color{red}{-\frac{8}{3}\cdot\ln(4)}\cancel{-\frac{8}{3}\cdot\ln(x)}+\left(4x-2x^2+\frac{x^3}{3}\right)\cdot \ln(4x)\cancel{+\frac{8}{3}\ln{(x)}}-4x+x^2-\frac{1}{9}x^3+\color{red}C=\\ \left(4x-2x^2+\frac{x^3}{3}\right)\cdot \ln(4x)-4x+x^2-\frac{1}{9}x^3+C.$$ Note: It would be shorter if it was used: $$dv=(2-x)^2dx \Rightarrow dv=(4-4x+x^2)dx \Rightarrow v=4x-2x^2+\frac{x^3}{3}.$$

Answer 2

$$\int(2-x)^2\ln{(4x)}dx$$

Integrate by parts

$$f = \ln{(4x)},\quad g′= (x- 2)^2 $$ $$f′=\frac{1}{x},\quad g=\frac{x^3}{3}−2x^2+4x$$ This follows $$\left(\frac{x^3}{3}−2x^2+4x\right) \ln(4x)− \underbrace{\int\frac{\frac{x^3}{3}−2x^2+4x}{x}\, dx}_{I}$$

\begin{align} I &= \int\left(\frac{x^2}{3}−2x+4\right)dx\\ &= \int\frac{x^2}{3}\,dx−2\int x\,dx+ \int4\,dx\\ &=\frac{x^3}{9}−x^2+4x \end{align}

So

$$\left(\frac{x^3}{3}−2x^2+4x\right) \ln(4x)− \frac{x^3}{9}+x^2-4x + C$$

Answer 3

dont substitute.

$\int(4-4x + x^2)\ln(4x)dx = (4x-2x^2+\frac{1}{3}x^3)\ln(4x)-\int\frac{(4x-2x^2+\frac{1}{3}x^3)}{x}dx$

$=(4x-2x^2+\frac{1}{3}x^3)\ln(4x) - \int(4-2x+\frac{1}{3}x^2)dx$

you can take it from here.