Evaluate the integral: I = integral x sqrt(x²-2x+2) dx
Evaluate the integral: $$ I = \int x\sqrt{x^2-2x+2}\,dx. $$
Explanation of Method
Completing the square gives $x^2-2x+2=(x-1)^2+1$; shifting to $u=x-1$ splits the integrand into $u\sqrt{u^2+1}$ (a direct power-rule term) plus $\sqrt{u^2+1}$ (the standard radical integral, via $\sec^3\theta$).
- Complete the square and shift $u=x-1$.
- Split into two integrals.
- Integrate each piece.
- Back-substitute.
Step 1 · Shift
Let $u=x-1$, so $x^2-2x+2=u^2+1$ and $x=u+1$:
$$ I = \int(u+1)\sqrt{u^2+1}\,du = \int u\sqrt{u^2+1}\,du + \int\sqrt{u^2+1}\,du. $$
Step 2 · First piece
$$ \int u\sqrt{u^2+1}\,du = \frac13(u^2+1)^{3/2} + C_1. $$
Step 3 · Second piece
Using the standard formula $\int\sqrt{u^2+1}\,du=\tfrac u2\sqrt{u^2+1}+\tfrac12\ln\left|u+\sqrt{u^2+1}\right|$:
$$ \int\sqrt{u^2+1}\,du = \frac{u\sqrt{u^2+1}}2 + \frac12\ln\left|u+\sqrt{u^2+1}\right| + C_2. $$
Step 4 · Combine and back-substitute
With $u=x-1$, $u^2+1=x^2-2x+2$, and combining the two $\sqrt{u^2+1}$ terms into a single polynomial factor:
$$ I = \frac{(2x^2-x+1)\sqrt{x^2-2x+2}}6 + \frac12\ln\left|x-1+\sqrt{x^2-2x+2}\right| + C. $$
Summary Table
| Piece | Antiderivative |
|---|---|
| $u\sqrt{u^2+1}$ | $\tfrac13(u^2+1)^{3/2}$ |
| $\sqrt{u^2+1}$ | $\tfrac u2\sqrt{u^2+1}+\tfrac12\ln|u+\sqrt{u^2+1}|$ |
Additional Notes
- The domain is all real $x$, since $x^2-2x+2=(x-1)^2+1>0$ everywhere.
- The result was verified by differentiation at $x=0$ and $x=2$: the derivative matches the integrand $x\sqrt{x^2-2x+2}$ exactly at both points.
Result
The solution to the integral is: