# How to find the integral of $\frac {1}{x^2-x+2}$? [duplicate]

I am learning homogeneous differential equations, and I tried to answer one of the questions, but I am stuck at integrating the equation above. Are there any easier ways than making the denominator: $$(x-\frac{x}{2})^2+\frac{7}{4}$$?
Hint. One may just observe that, by the chain rule, $$\left(\frac1b\arctan \frac{x-a}{b} \right)'_x=\frac1{(x-a)^2+b^2}.$$
Use that $$x^2-x+2=\left(x-\frac{1}{2}\right)^2+\frac{7}{8}$$ and substitute $$t=x-\frac{1}{2}$$