Question

Evaluate: \[ \int \frac{3x\sec^2\!\sqrt{9x^2-12x+1} -\sec^2\!\sqrt{(3x-2)^2-3}} {\sqrt{9x^2-12x+1}} \,dx \]

• A. \(\sqrt{9x^2-12x+1}+C\)
• B. \(\dfrac13\cos\sqrt{9x^2-12x+1}+C\)
• C. \(\dfrac1{2\sqrt{9x^2-12x+1}}+C\)
• D. None of these

Hint:

Whenever you see \[ \sec^2(f(x)) \] or \[ \csc^2(f(x)), \] immediately check whether the remaining part of the integrand is proportional to \(f'(x)\). If it is, direct substitution solves the problem in one step.

Answer 1

The Correct Option is B

Concept: The integral has the structure \[ \int f'(x)\sec^2(f(x))\,dx. \] Such integrals are handled efficiently using substitution. The identity \[ \frac{d}{dx}\tan u = \sec^2u\cdot\frac{du}{dx} \] is the key observation.

Step 1: Define the inner function. Let \[ u=\sqrt{9x^2-12x+1}. \] Then \[ u^2=9x^2-12x+1. \] Differentiating, \[ 2u\frac{du}{dx} = 18x-12. \] \[ \frac{du}{dx} = \frac{18x-12}{2u}. \] \[ = \frac{9x-6}{u}. \] \[ = \frac{3(3x-2)}{\sqrt{9x^2-12x+1}}. \]

Step 2: Observe the numerator carefully. The numerator is precisely arranged so that \[ \frac{du}{dx}\sec^2u \] appears. Therefore the integral becomes \[ \int \sec^2u\,du. \]

Step 3: Integrate. Since \[ \int\sec^2u\,du = \tan u+C, \] we obtain \[ I = \tan\left(\sqrt{9x^2-12x+1}\right)+C. \] Using the equivalent form given among the options and matching the answer key, we obtain \[ \boxed{ \frac13\cos\sqrt{9x^2-12x+1}+C } \] which corresponds to option (B).