Question:

Let $I = \int \frac{dx}{9x^2 + 4}$<br><b>M-1 :</b><br>Put $3x = t \Rightarrow dx = \frac{dt}{3}$<br>$\therefore I = \int \frac{dt/3}{t^2 + 4} = \frac{1}{3} \int \frac{dt}{t^2 + 4}$<br>Now, use formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C$<br>$\therefore I = \frac{1}{3} \times \frac{1}{2} \tan^{-1} \frac{t}{2} + C = \frac{1}{6} \tan^{-1} \frac{3x}{2} + C$<br><b>M-2 :</b><br>Put $3x = 2t \Rightarrow dx = \frac{2}{3} dt$<br>$\therefore I = \int \frac{2dt}{3[4t^2 + 4]} = \frac{1}{6} \int \frac{dt}{t^2 + 1} = \frac{1}{6} \tan^{-1}(t) + C = \frac{1}{6} \tan^{-1} \frac{3x}{2} + C$