Question

Evaluate the integral: \[ \int \frac{(3x - 2)\tan\left(\sqrt{9x^2 - 12x + 1}\right)}{\sqrt{9x^2 - 12x + 1}} \, dx =\ ?\]

• A. \(\frac{1}{3} \sec^2 \left( \sqrt{9x^2 - 12x + 1} \right) + c\)
• B. \(\frac{1}{3} \sec^2 x + c\)
• C. \(\frac{1}{2} \log \left| \sec \left( \sqrt{9x^2 - 12x + 1} \right) \right| + c\)
• D. \(\frac{1}{3} \log \left| \sec \left( \sqrt{9x^2 - 12x + 1} \right) \right| + c\)

Hint:

Look for expressions in the form \(\frac{f'(x)}{f(x)}\) or recognize the derivative of composite functions like \(\tan(\sqrt{g(x)})\) — this often points to using substitution and standard integrals involving \(\tan x\).

Answer 1

The Correct Option is D

Step 1: Let us denote: 
\[ u = \sqrt{9x^2 - 12x + 1} \Rightarrow u^2 = 9x^2 - 12x + 1 \] Step 2: Differentiate both sides to find \(\frac{du}{dx}\): 
\[ 2u \cdot \frac{du}{dx} = 18x - 12 \Rightarrow \frac{du}{dx} = \frac{18x - 12}{2u} = \frac{6(3x - 2)}{u} \] Step 3: Now examine the integrand: 
\[ \int \frac{(3x - 2)\tan(u)}{u} \, dx \] Use the substitution from Step 2: 
\[ dx = \frac{u}{6(3x - 2)} \, du \Rightarrow \int \frac{(3x - 2)\tan(u)}{u} \cdot \frac{u}{6(3x - 2)} \, du = \frac{1}{6} \int \tan(u) \, du \] Step 4: Integrate: 
\[ \int \tan(u) \, du = \log |\sec(u)| + c \Rightarrow \frac{1}{6} \log |\sec(u)| + c \] Step 5: Recall \(u = \sqrt{9x^2 - 12x + 1}\), so: 
\[ \frac{1}{6} \log \left| \sec\left(\sqrt{9x^2 - 12x + 1} \right) \right| + c \] But we were integrating: 
\[ \int \frac{(3x - 2)\tan(u)}{u} \, dx \] We substituted and got: 
\[ = \frac{1}{6} \cdot 2 \log \left| \sec(u) \right| + c = \frac{1}{3} \log \left| \sec\left( \sqrt{9x^2 - 12x + 1} \right) \right| + c \]