Inte(grity)

We have:

$\large{\int\dfrac{6x^{4}-5x^{3}+4x^{2}}{2x^{2}-x+1}dx}$

$=\large{\int\dfrac{x^{2}(3(2x^{2}-x+1)-2x+1)}{2x^{2}-x+1}dx}$

$=\large{\int 3x^{2}dx-\int\dfrac{x(2x^{2}-x+1-1)}{2x^{2}-x+1}dx}$

$=\large{\int 3x^{2}dx-\int xdx +\frac{1}{4}\int\dfrac{4x-1+1}{2x^{2}-x+1}dx}$

$=\large{\int 3x^{2}dx -\int xdx +\frac{1}{4}[\int\dfrac{4x-1}{2x^{2}-x+1}dx+\int\dfrac{2}{(2x-\frac{1}{2})^{2}+(\frac{\sqrt{7}}{2})^{2}}dx]}$

Now:

$\large{\int\dfrac{4x-1}{2x^{2}-x+1}dx=\int\dfrac{d(2x^{2}-x+1)}{2x^{2}-x+1}=ln(|2x^{2}-x+1|)}$

also,

$\large{\int\dfrac{2}{(2x-\frac{1}{2})^{2}+(\frac{\sqrt{7}}{2})^{2}}dx=\int\dfrac{d(2x-\frac{1}{2})}{(2x-\frac{1}{2})^{2}+(\frac{\sqrt{7}}{2})^{2}}=\frac{2}{\sqrt{7}}tan^{-1}\dfrac{4x-1}{\sqrt{7}}}$

$\therefore \text{ we get }$ $\large{1.x^{3}-\frac{x^{2}}{2}+\frac{1}{4}ln(|2x^{2}-x+1|)+\frac{1}{2\sqrt{7}}tan^{-1}\dfrac{4x-1}{\sqrt{7}}}$

comparing, we get:

$a_1=1;a_2=2;a_3=4;a_4=2;a_5=2;a_6=7;a_7=4;a_8=7$

our sum: $\boxed{29}$