Evaluate the integral: I = integral dx/(3+x-x²)

Evaluate the integral: $$ I = \int \frac{dx}{3+x-x^2}. $$

Step 1 · Complete the square

$$ 3+x-x^2 = -\left(x^2-x-3\right) = -\left[\left(x-\frac12\right)^2 - \frac14 - 3\right] = \frac{13}{4} - \left(x-\frac12\right)^2. $$

Step 2 · Substitute

Let $u=x-\tfrac12$, $du=dx$, and $a=\dfrac{\sqrt{13}}{2}$:

$$ I = \int \frac{du}{a^2-u^2}. $$

Step 3 · Apply the standard formula

$$ I = \frac{1}{2a}\ln\left|\frac{a+u}{a-u}\right| + C = \frac{1}{\sqrt{13}}\ln\left|\frac{\sqrt{13}/2+u}{\sqrt{13}/2-u}\right| + C. $$

Step 4 · Substitute back

With $u=x-\tfrac12$, multiply numerator and denominator inside the log by $2$:

$$ I = \frac{1}{\sqrt{13}}\ln\left|\frac{\sqrt{13}+2x-1}{\sqrt{13}-2x+1}\right| + C = \frac{\sqrt{13}}{13}\ln\left|\frac{2x-1+\sqrt{13}}{1-2x+\sqrt{13}}\right| + C. $$

Summary Table

Additional Notes

• Correction: the source claimed $x^2-x-3=(x-3)(x+2)$, but expanding $(x-3)(x+2)=x^2-x-6$, not $x^2-x-3$ — the constant term is wrong. Since the discriminant of $x^2-x-3$ is $1+12=13$, which is not a perfect square, this quadratic has irrational roots and does not factor over the rationals at all; the whole partial-fraction approach in the source is invalid. Completing the square and using the $\int du/(a^2-u^2)$ formula is the correct route.
• The domain requires $3+x-x^2\neq0$, i.e. $x\neq\frac{1\pm\sqrt{13}}{2}$.
• The result can be checked by differentiating $\tfrac{1}{\sqrt{13}}\ln\left|\tfrac{\sqrt{13}+2x-1}{\sqrt{13}-2x+1}\right|$: writing it as $\tfrac{1}{\sqrt{13}}\left[\ln|\sqrt{13}+2x-1|-\ln|\sqrt{13}-2x+1|\right]$, the derivative is $\tfrac{1}{\sqrt{13}}\left[\tfrac{2}{\sqrt{13}+2x-1}+\tfrac{2}{\sqrt{13}-2x+1}\right]=\tfrac{2}{\sqrt{13}}\cdot\tfrac{2\sqrt{13}}{13-(2x-1)^2}=\tfrac{4}{13-(2x-1)^2}=\tfrac{4}{4(3+x-x^2)}=\tfrac{1}{3+x-x^2}$, matching the integrand.