Integrál

$=\int \frac{x^2-2x+1}{\sqrt{x}}dx=$ $\int x^{\frac{3}{2}}-2x^{\frac{1}{2}}+x^{-\frac{1}{2}}dx=$ $\frac{2}{5}{x}^{\frac{5}{2}}-\frac{4}{3}{x}^{\frac{3}{2}}-2x^{\frac{1}{2}}+c$

$=\int x^{-\frac{1}{4}}dx$ $=\frac{4}{3}{x}^{\frac{3}{4}}+c$

$=\int \frac{2sinxcosx}{cosx}dx$ $=\int 2sinxdx$ $=-2cosx+c$

$=\left|\begin{array}{c}u=5x-1\\ \frac{1}{5}du=dx\end{array}\right|$ $=\int \frac{1}{5}{u}^{3}du$ $=\frac{1}{20}{u}^4+c$ $=\frac{1}{20}{(5x-1)}^4+c$

$=\left|\begin{array}{c}u=lnx\\ du=\frac{1}{x}dx\end{array}\right|$ $=\int \frac{1}{\sqrt{u}}du$
$=\int u^{-\frac{1}{2}}du$
$=2u^{\frac{1}{2}}+c$ $=2{(lnx)}^{\frac{1}{2}}+c$

$=\left|\begin{array}{c}u=2x\\ \frac{1}{2}du=dx\end{array}\right|$ $=\int \frac{1}{2}sinudu$
$=-\frac{1}{2}cosu+c$
$=-\frac{1}{2}cos2x+c$

$=14x+8$

$=\frac{2(2x+1)-2(2x-1)}{{(2x+1)}^2}$ $=\frac{4}{{(2x+1)}^2}$