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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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.

Similar Questions

$\int {\frac{{2x + 5}}{{\sqrt {7 - 6x - {x^2}} }}dx} = A\sqrt {7 - 6x - {x^2}} + B\,{\sin ^{ - 1}}\left( {\frac{{x + 3}}{4}} \right) + C$ (where $C$ is a constant of integration),then the ordered pair $(A, B)$ is equal to

Integrate the function: $\int \sqrt{x^{2}+4 x+6} \, dx$

$ \int \frac{1}{x^{2}\left(x^{4}+1\right)^{3 / 4}} dx $ is equal to

If $\int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x \sin (x-\theta)}} d x=A \sqrt{\cos \theta \tan x-\sin \theta}+B \sqrt{\cos \theta-\cot x \sin \theta}+C,$ where $C$ is the integration constant,then $AB$ is equal to

Let $f(x) = \int \frac{dx}{(3+4x^2) \sqrt{4-3x^2}}$,$|x| < \frac{2}{\sqrt{3}}$. If $f(0) = 0$ and $f(1) = \frac{1}{\alpha \beta} \tan^{-1}\left(\frac{\alpha}{\beta}\right)$,where $\alpha, \beta > 0$,then $\alpha^2 + \beta^2$ is equal to $.........$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module