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

If $I_1 = \int \sin^6 x \, dx$ and $I_2 = \int \cos^6 x \, dx$, then $I_1 + I_2 = $

Let $I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}}$. If $I(37) - I(24) = \frac{1}{4} \left( \frac{1}{b^{\frac{1}{13}}} - \frac{1}{c^{\frac{1}{13}}} \right)$,where $b, c \in \mathbb{N}$,then $3(b+c)$ is equal to

$\int \frac{x-1}{(x+1) \sqrt{x^3+x^2+x}} d x=$

Let $\int \frac{2-\tan x}{3+\tan x} dx = \frac{1}{2}(\alpha x + \log_e |\beta \sin x + \gamma \cos x|) + C$,where $C$ is the constant of integration. Then $\alpha + \frac{\gamma}{\beta}$ is equal to:

If $\int \frac{1-(\cot x)^{2019}}{\tan x+(\cot x)^{2020}} dx = \frac{1}{n} \ln |(f(x))^n + (g(x))^n| + c$, then the value of $n[(f(x))^4 + (g(x))^4]_{x=\frac{\pi}{3}}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module