Answer
$\left(x-\frac{1}{2}\right)\ln(x^2-x+2)-2x+{\sqrt 7}\tan^{-1}\left(\frac{2x-1}{\sqrt 7}\right)+C$
Work Step by Step
$u$ = $\ln(x^2-x+2)$
$du$ = $\frac{2x-1}{x^2-x+2}dx$
Integrate by parts:
$\int{\ln(x^2-x+2)}dx$ = $x\ln(x^2-x+2)-\int\frac{2x^2-x}{x^2-x+2}dx$
= $x\ln(x^2-x+2)-\int\left(2+\frac{x-4}{x^2-x+2}\right)dx$
= $x\ln(x^2-x+2)-2x-\int\left(\frac{\frac{1}{2}(2x-1)}{x^2-x+2}\right)dx+\frac{7}{2}\int\left(\frac{1}{(x-\frac{1}{2})^2+\frac{7}{4}}\right)dx$
= $x\ln(x^2-x+2)-2x-\frac{1}{2}\ln(x^2-x+2)+{\sqrt 7}\tan^{-1}\left(\frac{2x-1}{\sqrt 7}\right)+C$
= $\left(x-\frac{1}{2}\right)\ln(x^2-x+2)-2x+{\sqrt 7}\tan^{-1}\left(\frac{2x-1}{\sqrt 7}\right)+C$
