Question Detail

This integral is of the form (\int \frac{du}{a^2+u^2} = \frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right) + C).Here, (a^2 = 4 \implies a = 2).And (u^2 = 9x^2 \implies u = 3x).Then (du = 3\,dx \implies 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})(= \frac{1}{3} \cdot \frac{1}{2} \tan^{-1}\left(\frac{u}{2}\right) + C)(= \frac{1}{6} \tan^{-1}\left(\frac{3x}{2}\right) + C).