Question

Evaluate: \[ \int \sqrt{x^2 - 8x + 7}\, dx \]

Hint:

For integrals of the form $\sqrt{x^2+bx+c}$, first complete the square, then apply standard formula for $\sqrt{u^2-a^2}$.

Answer 1

Step 1: Simplify quadratic.
\[ x^2 - 8x + 7 = (x^2 - 8x + 16) - 9 = (x-4)^2 - 3^2. \] So integral becomes: \[ I = \int \sqrt{(x-4)^2 - 3^2}\, dx. \]

Step 2: Use standard formula.
Recall: \[ \int \sqrt{u^2 - a^2}\, du = \frac{u}{2}\sqrt{u^2 - a^2} - \frac{a^2}{2}\ln \left| u + \sqrt{u^2 - a^2}\right| + C. \] Here $u = (x-4)$ and $a=3$, so $du=dx$.

Step 3: Apply formula.
\[ I = \frac{(x-4)}{2}\sqrt{(x-4)^2 - 9} - \frac{9}{2}\ln \left| (x-4) + \sqrt{(x-4)^2 - 9}\right| + C. \]

Final Answer: \[ \boxed{ \frac{(x-4)}{2}\sqrt{(x-4)^2 - 9} - \frac{9}{2}\ln \left| (x-4) + \sqrt{(x-4)^2 - 9}\right| + C } \]