Evaluate the integral: I = integral dx/((4-2x+x²)
Evaluate the integral: $$ I = \int \frac{dx}{\left(4-2x+x^2\right)\sqrt{2+2x-x^2}}. $$
Explanation of Method
Completing the square in both quadratics and shifting $u=x-1$ turns the problem into $\int du/[(u^2+3)\sqrt{3-u^2}]$; the substitution $u=\sqrt3\sin\theta$ then reduces this to a rational-in-$\tan\theta$ integral.
- Complete the square in both quadratics and shift $u=x-1$.
- Substitute $u=\sqrt3\sin\theta$.
- Convert to tangent form and integrate.
- Back-substitute.
Step 1 · Complete the square and shift
$4-2x+x^2=(x-1)^2+3$ and $2+2x-x^2=3-(x-1)^2$. Let $u=x-1$:
$$ I = \int\frac{du}{\left(u^2+3\right)\sqrt{3-u^2}}. $$
Step 2 · Trig substitution
Let $u=\sqrt3\sin\theta$, so $du=\sqrt3\cos\theta\,d\theta$, $\sqrt{3-u^2}=\sqrt3\cos\theta$, $u^2+3=3(1+\sin^2\theta)$:
$$ I = \int\frac{\sqrt3\cos\theta\,d\theta}{3(1+\sin^2\theta)\cdot\sqrt3\cos\theta} = \frac13\int\frac{d\theta}{1+\sin^2\theta}. $$
Step 3 · Convert to tangent form
Dividing by $\cos^2\theta$ and using $\sec^2\theta=1+\tan^2\theta$:
$$ I = \frac13\int\frac{\sec^2\theta\,d\theta}{1+2\tan^2\theta}. $$
Let $t=\tan\theta$:
$$ I = \frac13\int\frac{dt}{1+2t^2} = \frac1{3\sqrt2}\arctan\left(\sqrt2t\right) + C. $$
Step 4 · Back-substitute
With $t=\tan\theta=u/\sqrt{3-u^2}$ and $u=x-1$, $3-u^2=2+2x-x^2$:
$$ I = \frac1{3\sqrt2}\arctan\left(\frac{\sqrt2(x-1)}{\sqrt{2+2x-x^2}}\right) + C. $$
Summary Table
| Substitution | Effect |
|---|---|
| $u=x-1$ | reduces to $\int du/[(u^2+3)\sqrt{3-u^2}]$ |
| $u=\sqrt3\sin\theta$ | reduces to $\tfrac13\int d\theta/(1+\sin^2\theta)$ |
Additional Notes
- The domain requires $2+2x-x^2>0$, i.e. $1-\sqrt3<x<1+\sqrt3$.
- The result was verified by differentiation at $x=0$ and $x=1$: the derivative matches the integrand $1/[(4-2x+x^2)\sqrt{2+2x-x^2}]$ exactly at both points.
Result
The solution to the integral is: