In this substitution, I tried the method of complete the square, but at the end the answer is not correct. I'm totally sure that the answer is: $\frac{2\sqrt{7}}{7}\arctan(\frac{\sqrt{7}(2x-1)}{7})+C$, and I try different techniques, but I can´t get this result.

This is the Integral:


  • $\begingroup$ Look at my generalised answer here and check $\endgroup$ – Oussama Boussif Aug 26 '15 at 15:23

\begin{align*} \int\frac{dx}{x^2-x+2}&=\int\frac{dx}{\left(x-\frac{1}{2}\right)^2+\frac{7}{4}}\\ &=\frac{2}{\sqrt{7}}\arctan\left(\frac{2x-1}{\sqrt{7}}\right)+C \end{align*}

Since $$\int\frac{du}{u^2+a^2}=\frac{1}{a}\arctan\left(\frac{u}{a}\right)+C$$


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