How do I integrate (sqrt(x-2))(x^{2 }+ x + 1)
I've got a feeling it's some basic method of integration that I'm forgetting, but I can't find anything like it in my notes.
Start by writing
$\displaystyle \displaystyle \begin{align*} t &= x - 2 \\ x &= t + 2 \\ x^2 &= (t + 2)^2 \\ x^2 &= t^2 + 4t + 4 \\ x^2 + x &= t^2 + 4t + 4 + t + 2 \\ x^2 + x &= t^2 + 5t + 6 \\ x^2 + x + 1 &= t^2 + 5t + 7 \\ x^2 + x + 1 &= (x - 2)^2 + 5(x - 2) + 7 \end{align*}$
Now rewrite the integrand as $\displaystyle \displaystyle \begin{align*} \frac{\sqrt{x - 2}}{x^2 + x + 1} = \frac{2(x - 2)}{2\sqrt{x - 2}\left[(x - 2)^2 + 5(x - 2) + 7\right] } \end{align*}$
Now if you make the substitution $\displaystyle \displaystyle \begin{align*} u = \sqrt{x - 2} \implies du = \frac{1}{2\sqrt{x - 2}}\,dx \end{align*}$ the integral becomes
$\displaystyle \displaystyle \begin{align*} \int{\frac{2u^2}{u^4 + 5u^2 + 7}\,du} \end{align*}$
Can you go from here?