Evaluate the integral: I = integral dx/(x²-x+2)

Evaluate the integral: $$ I = \int \frac{dx}{x^2-x+2}. $$

Step 1 · Complete the square

$$ x^2-x+2 = \left(x-\frac12\right)^2 + \frac74. $$

Step 2 · Substitute

Let $u=x-\tfrac12$, $du=dx$:

$$ I = \int \frac{du}{u^2+\tfrac74} = \int \frac{du}{u^2+\left(\tfrac{\sqrt7}{2}\right)^2}. $$

Step 3 · Apply the standard arctangent formula

With $a=\dfrac{\sqrt7}{2}$:

$$ I = \frac{1}{a}\arctan\frac{u}{a} + C = \frac{2}{\sqrt7}\arctan\left(\frac{2u}{\sqrt7}\right) + C. $$

Step 4 · Substitute back

With $u=x-\tfrac12$, so $2u=2x-1$:

$$ I = \frac{2}{\sqrt7}\arctan\left(\frac{2x-1}{\sqrt7}\right) + C = \frac{2\sqrt7}{7}\arctan\left(\frac{2x-1}{\sqrt7}\right) + C. $$

Summary Table

Additional Notes

• Correction: as in Problem 189, the source's intermediate "factor out $1/4$" step double-counted a constant, producing $\tfrac{8}{7\sqrt7}$ instead of the correct $\tfrac{2}{\sqrt7}$. Applying $\int du/(u^2+a^2)=(1/a)\arctan(u/a)$ directly with $a=\sqrt7/2$ gives the coefficient $2/\sqrt7=2\sqrt7/7$.
• The integrand is defined for all real $x$, since $x^2-x+2>0$ everywhere (its discriminant $1-8=-7$ is negative).
• The result can be checked by differentiating $\tfrac{2}{\sqrt7}\arctan\!\left(\tfrac{2x-1}{\sqrt7}\right)$: this gives $\tfrac{2}{\sqrt7}\cdot\tfrac{2/\sqrt7}{1+(2x-1)^2/7}=\tfrac{4/7}{[7+(2x-1)^2]/7}=\tfrac{4}{7+(2x-1)^2}=\tfrac{4}{4x^2-4x+8}=\tfrac{1}{x^2-x+2}$, matching the integrand.