I have the following problem:

Given the parabola y=x^2-4x+3 find the area and the parimeter of OAB

A=(1,0), B=(0,3), C=(3,0)

I calculated the area

\(\displaystyle

S oab= \int ^{1}_{0}\left( x^{2}-4x+3\right) dx

=\left| \left( \dfrac {x^{3}}{3}-\dfrac {4x^{2}}{2}+3x\right) \right|

\ = \left| \left( \dfrac {1}{3}-\dfrac {4}{3}+3\right) -0\right| \ = \dfrac {4}{3}

\)

However i am stuck into the parimeter.

\(\displaystyle

P oab=\left| OA\right| +\left| OB\right| +\left| AB\right|\\

= 1 +3 + \int ^{1}_{0}\sqrt {1+\left( \dfrac {d}{dy}\right) ^{2}dx}\\

\begin{aligned}=4+\int ^{1}_{0}\sqrt {1+\left( 2x-4\right) ^{2}}d_{x}\\

=4+\int ^{1}_{0}\sqrt {4x-16x+12}dx\\ =4+2\int ^{1}_{0}\sqrt {x^{2}-4x+\dfrac {17}{4}}dx\\

=4+2\int ^{1}_{0}\sqrt {\left( x-2\right) ^{2}+\dfrac {1}{4}}dx\end{aligned}

\)

Here i tried to substitute u= x-2, but i couldnt solve it.