int sqrt{x^2-6x-16} , dx equals. | vedclass

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(B) To evaluate the integral $I = \int \sqrt{x^2-6x-16} \, dx$,we first complete the square for the quadratic expression inside the square root: $x^2-

Vedclass · Accepted Answer

$\left(\frac{x-3}{2}\right) \sqrt{x^2-6x-16} - \frac{25}{2} \log \left|x-3+\sqrt{x^2-6x-16}\right| + c$

Vedclass · Answer

(B) To evaluate the integral $I = \int \sqrt{x^2-6x-16} \, dx$,we first complete the square for the quadratic expression inside the square root: $x^2-6x-16 = (x^2-6x+9) - 9 - 16 = (x-3)^2 - 25 = (x-3)^2 - 5^2$. Now,the integral becomes $I = \int \sqrt{(x-3)^2 - 5^2} \, dx$. Using the standard integration formula $\int \sqrt{t^2-a^2} \, dt = \frac{t}{2} \sqrt{t^2-a^2} - \frac{a^2}{2} \log |t + \sqrt{t^2-a^2}| + c$,where $t = x-3$ and $a = 5$: $I = \frac{x-3}{2} \sqrt{(x-3)^2 - 5^2} - \frac{5^2}{2} \log |(x-3) + \sqrt{(x-3)^2 - 5^2}| + c$. Substituting back the original expression $x^2-6x-16$: $I = \frac{x-3}{2} \sqrt{x^2-6x-16} - \frac{25}{2} \log |x-3 + \sqrt{x^2-6x-16}| + c$.

$\int \sqrt{x^2-6x-16} \, dx$ equals.

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