Answer
$$
\int \ln \left(x^{2}-x+2\right) d x = \left(x-\frac{1}{2}\right) \ln \left(x^{2}-x+2\right)-2 x+\sqrt{7} \tan ^{-1} \frac{2 x-1}{\sqrt{7}}+C
$$
where $C$ be an arbitrary constant.
Work Step by Step
$$
\int \ln \left(x^{2}-x+2\right) d x
$$
$$
\text { Let } u=\ln \left( x^{2}-x+2\right), \quad\quad d v=d x .\\
\text { Then } d u=\frac{2 x-1}{x^{2}-x+2} d x, v=x,
$$
and (by integration by parts) , we have :
\begin{aligned}
\int \ln \left(x^{2}-x+2\right) d x &=x \ln \left(x^{2}-x+2\right)-\int \frac{2 x^{2}-x}{x^{2}-x+2} d x \\
& =x \ln \left(x^{2}-x+2\right)-\int\left(2+\frac{x-4}{x^{2}-x+2}\right) d x \\
&=x \ln \left(x^{2}-x+2\right)-2 x-\int \frac{\frac{1}{2}(2 x-1)}{x^{2}-x+2} d x + \\
& \quad\quad \quad\quad +\frac{7}{2} \int \frac{d x}{\left(x-\frac{1}{2}\right)^{2}+\frac{7}{4}} \\
&=x \ln \left(x^{2}-x+2\right)-2 x-\frac{1}{2} \ln \left(x^{2}-x+2\right)+ \\
& \quad\quad \quad\quad +\frac{7}{2} \int \frac{\frac{\sqrt{7}}{2} d u} {\frac{7}{4}\left(u^{2}+1\right)} \\
&\quad\quad \quad\quad \quad\quad \quad\left[\begin{array}{c}
\text { where } x-\frac{1}{2}=\frac{\sqrt{7}}{2} u, \\
d x=\frac{\sqrt{7}}{2} d u, \\
\left(x-\frac{1}{2}\right)^{2}+\frac{7}{4}=\frac{7}{4}\left(u^{2}+1\right)
\end{array}\right] \\
&=\left(x-\frac{1}{2}\right) \ln \left(x^{2}-x+2\right)-2 x+\sqrt{7} \tan ^{-1} u+C \\
&=\left(x-\frac{1}{2}\right) \ln \left(x^{2}-x+2\right)-2 x+\sqrt{7} \tan ^{-1} \frac{2 x-1}{\sqrt{7}}+C
\end{aligned}
So the integral:
$$
\int \ln \left(x^{2}-x+2\right) d x = \left(x-\frac{1}{2}\right) \ln \left(x^{2}-x+2\right)-2 x+\sqrt{7} \tan ^{-1} \frac{2 x-1}{\sqrt{7}}+C
$$
where $C$ be an arbitrary constant.
