A physics problem says that the following integral is said to evaluate to: $$\int_{}{} \frac{x^2}{\sqrt{x^2-x+1}} dx = \frac{2x+3}{8}\sqrt{x^2-x+1} + \frac{1}{16}\ln(2x+1+2\sqrt{x^2-x+1})$$
I can not seem to find a way that evaluates to this. I know that a integral of this form can be broken down to:
$$\int_{}{} \frac{P(x)}{\sqrt{ax^2+bx+c}}dx = Q(x)\sqrt{ax^2+bx+c} + t\int{}{}\frac{1}{\sqrt{ax^2+bx+c}}dx$$
Where $Q(x)$ is a polynomial of degree one less than $P(x)$. Can anyone see a solution from this angle?
Equate numerator as $x^2=A(x^2-x+1)+B(2x-1)+C$
which gives $A=1, B=\frac12$ & $C=-\frac12$. Now proceed as follows $$\int_{}{} \frac{x^2}{\sqrt{x^2-x+1}} dx$$ $$=\int \frac{(x^2-x+1)+\frac12(2x-1)-\frac12}{\sqrt{x^2-x+1}} dx$$ $$=\int\left( \sqrt{x^2-x+1} +\frac12\frac{(2x-1)}{\sqrt{x^2-x+1}}-\frac12\frac{1}{\sqrt{x^2-x+1}} \right)dx$$ $$=\int \sqrt{\left(x-\frac12\right)^2+\frac34}\ dx +\frac12\int\frac{d(x^2-x+1)}{\sqrt{x^2-x+1}} -\frac12\int \frac{dx}{\sqrt{\left(x-\frac12\right)^2+\frac34}}$$