Question Detail

This integral is of the form $\int \frac{du}{a^2+u^2} = \frac{1}{a} \tan^{-1}(\frac{u}{a}) + C$.In our integral, $4 = 2^2$, so $a=2$.For $9x^2$, let $u^2 = 9x^2$, so $u = 3x$.Then, $du = 3 dx$, which means $dx = \frac{1}{3} du$.Substitute these into the integral:$\int \frac{\frac{1}{3} du}{2^2+u^2} = \frac{1}{3} \int \frac{du}{2^2+u^2}$Now apply the formula:$= \frac{1}{3} \cdot \frac{1}{2} \tan^{-1}(\frac{u}{2}) + C$$= \frac{1}{6} \tan^{-1}(\frac{3x}{2}) + C$.Thus, option B is the correct answer.