Evaluate the integral: I = integral

Evaluate the integral: $$ I = \int \frac{1-x+x^2}{\sqrt{1+x-x^2}}\,dx. $$

Step 1 · Rewrite the numerator

Matching coefficients in $1-x+x^2=A(1+x-x^2)+B(1-2x)+C$ gives $A=-1,\ B=0,\ C=2$, so:

$$ \frac{1-x+x^2}{\sqrt{1+x-x^2}} = -\sqrt{1+x-x^2} + \frac2{\sqrt{1+x-x^2}}. $$

Step 2 · Complete the square and integrate

$1+x-x^2=\tfrac54-(x-\tfrac12)^2$. Using $\int dx/\sqrt{5/4-(x-1/2)^2}=\arcsin\left(\tfrac{2x-1}{\sqrt5}\right)$ and $\int\sqrt{5/4-(x-1/2)^2}\,dx=\tfrac{2x-1}4\sqrt{1+x-x^2}+\tfrac58\arcsin\left(\tfrac{2x-1}{\sqrt5}\right)$:

$$ I = -\left[\frac{2x-1}4\sqrt{1+x-x^2}+\frac58\arcsin\left(\frac{2x-1}{\sqrt5}\right)\right] + 2\arcsin\left(\frac{2x-1}{\sqrt5}\right) + C. $$

Step 3 · Combine

$$ I = -\frac{(2x-1)\sqrt{1+x-x^2}}4 + \left(2-\frac58\right)\arcsin\left(\frac{2x-1}{\sqrt5}\right) + C. $$

Summary Table

Additional Notes

• The domain is $\dfrac{1-\sqrt5}2<x<\dfrac{1+\sqrt5}2$.
• The result was verified numerically at $x=0$: the derivative evaluates to $1$, matching the integrand $(1-0+0)/\sqrt{1+0-0}=1$ there.