Question

use any method to evaluate the integral
$int \frac{x dx}{sqrt{121x^2 - 1}}$
$int \frac{x dx}{sqrt{121x^2 - 1}} = square$

Explanation:

Step1: Choose substitution

Let $u = 121x^2 - 1$.

Step2: Compute derivative of u

$\frac{du}{dx} = 242x \implies xdx = \frac{du}{242}$

Step3: Rewrite integral in terms of u

$\int \frac{x dx}{\sqrt{121x^2 - 1}} = \int \frac{1}{\sqrt{u}} \cdot \frac{du}{242}$

Step4: Simplify and integrate

$\frac{1}{242} \int u^{-1/2} du = \frac{1}{242} \cdot 2u^{1/2} + C = \frac{1}{121}u^{1/2} + C$

Step5: Substitute back u

Replace $u$ with $121x^2 - 1$

Answer:

$\frac{1}{121}\sqrt{121x^2 - 1} + C$