Evaluate the integral: I = integral sqrt(x²-2x-1) dx

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

Step 1 · Complete the square

$$ x^2-2x-1 = (x-1)^2-2. $$

Step 2 · Substitute

Let $u=x-1$, $du=dx$, $a^2=2$:

$$ I = \int \sqrt{u^2-2}\,du. $$

Step 3 · Apply the standard formula

$$ I = \frac u2\sqrt{u^2-2} - \ln\left|u+\sqrt{u^2-2}\right| + C. $$

Step 4 · Substitute back

With $u=x-1$:

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

Summary Table

Additional Notes

• Correction: the source offered two solutions that disagreed in sign on the logarithmic term. "Solution 1" (via the hyperbolic substitution $x-1=\sqrt2\sinh t$) picked up a sign error somewhere in the "after simplifications" step and produced $+\ln|\cdots|$; "Solution 2" (via the direct standard-formula route shown above) correctly produced $-\ln|\cdots|$. Differentiating the standard formula $\int\sqrt{u^2-a^2}\,du=\tfrac u2\sqrt{u^2-a^2}-\tfrac{a^2}2\ln|u+\sqrt{u^2-a^2}|$ confirms the minus sign is correct.
• The domain requires $x^2-2x-1\geq0$, i.e. $x\leq1-\sqrt2$ or $x\geq1+\sqrt2$.
• The result can be checked by differentiating $\tfrac{x-1}2\sqrt{x^2-2x-1}-\ln|x-1+\sqrt{x^2-2x-1}|$: the product-rule derivative of the first term is $\tfrac12\sqrt{x^2-2x-1}+\tfrac{(x-1)^2}{2\sqrt{x^2-2x-1}}$, and the derivative of the log term is $-\tfrac{1}{\sqrt{x^2-2x-1}}$; combining over the common denominator and using $(x-1)^2-2=x^2-2x-1$ reduces the sum to exactly $\sqrt{x^2-2x-1}$.