I'm trying to evaluate $\int\frac{3x}{\sqrt{1-2x}}dx$
This is what I got so far:
Let $u$ = $1-2x$
$x$ = $\frac{u-1}{-2}$
$du$ = $-2$ $dx$
$\frac{-du}{2}$ = $dx$
Therefore,
$\int\frac{3\frac{u-1}{-2}}{\sqrt{u}}\frac{-du}{2}$
$\int\frac{3(u-1)}{-2\sqrt{u}}\frac{-du}{2}$
$\frac{3}{4}\int\frac{(u-1)}{\sqrt{u}}{du}{}$
$\frac{3}{4}\int\frac{u}{{u^\frac{1}{2}}}-\frac{1}{{u^\frac{1}{2}}}{du}$
$\frac{3}{4}\int u^\frac{1}{2} -u^\frac{-1}{2}{du}$
$\frac{3}{4}[\frac{2}{3}u^\frac{3}{2} -2u^\frac{1}{2}]+C$
$\frac{6}{12}u^\frac{3}{2} -\frac{6}{4}u^\frac{1}{2}+C$
$\frac{1}{2}u^\frac{3}{2} -\frac{3}{2}u^\frac{1}{2}+C$
$\frac{1}{2}u^\frac{3}{2} -\frac{3}{2}u^\frac{1}{2}+C$
$\frac{u^\frac{3}{2}-3u^\frac{1}{2}}{2}+C$
$\frac{\sqrt{u^3}-3\sqrt{u}}{2}+C$
$\frac{\sqrt{(1-2x)^3}-3\sqrt{1-2x}}{2}+C$
The answer I'm supposed to get is $-\sqrt{1-2x} {(x+1)}+C$
The answer is $$\sqrt{(1-2x)^3}=\sqrt{(1-2x)(1-2x)^2}=\sqrt{(1-2x)}\sqrt{(1-2x)^2}=\color{red}{\sqrt{(1-2x)}}(1-2x) \\=\frac{\color{red}{\sqrt{(1-2x)}}(1-2x)-3\color{red}{\sqrt{1-2x}}}{2}+C \\=\sqrt{1-2x}[\frac{(1-2x)-3}{2}]+C \\=-\sqrt{1-2x}[x+1]+C$$ this is the correct answer, and your book has a typo.