Question

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. $$ \int \frac{d x}{3 x \sqrt{9 x^{2}-16}}=\frac{1}{4} \operatorname{arcsec} \frac{3 x}{4}+C $$

Short answer

The statement is false because the derivative of the right hand side does not equal the integrand on the left side of the equation.

Step-by-step solution

Compute The Derivative Of The Right Side Of The Equation

Let's compute the derivative of the right side of the equation and compare it to the given function inside the integral on the left side. The derivative of \(f(x) = \frac{1}{4} \operatorname{arcsec}\frac{3 x}{4}+C\) by using the chain rule and the knowledge that the derivative of \( \operatorname{arcsec} x\) is \( \frac{1}{|x|\sqrt{x^2-1}}\), we get: \(f'(x) = \frac{1}{4}*\frac{1}{|3x/4|\sqrt{(3x/4)^2-1}} * \frac{3}{4}\).

Simplifying The Result

Simplify this expression: \(f'(x) = \frac{3}{4|x|\sqrt{9x^2-16}}\). As |x| is just x (since x is under a square root and so it is always positive), the equation becomes: \(f'(x) = \frac{3}{4x\sqrt{9x^2-16}}\).

Comparing The Results

Now we can see that the derivative \(f'(x) = \frac{3}{4x\sqrt{9x^2-16}}\) is not equivalent to the integrand \( \frac{1}{3 x \sqrt{9 x^{2}-16}}\). This means that the given statement is false.