Question:

M-1 :<br>Let $I = \int \frac{dx}{2x^2+x-1} = \int \frac{dx}{(x+1)(2x-1)} = \frac{1}{3} \int \frac{2(x+1)-(2x-1)}{(x+1)(2x-1)} dx = \frac{1}{3} \int \left( \frac{2}{2x-1} - \frac{1}{x+1} \right) dx = \frac{1}{3} \ln \left| \frac{2x-1}{x+1} \right| + C<br>M-2 :<br>$I = \int \frac{dx}{2x^2+x-1} = \frac{1}{2} \int \frac{dx}{x^2+\frac{x}{2}-\frac{1}{2}} = \frac{1}{2} \int \frac{dx}{x^2+\frac{x}{2}+\frac{1}{16}-\frac{1}{16}-\frac{1}{2}} = \frac{1}{2} \int \frac{dx}{\left(x+\frac{1}{4}\right)^2 - \left(\frac{3}{4}\right)^2} = \frac{1}{2} \int \frac{dt}{t^2-a^2} \left[ \begin{matrix} t = x + \frac{1}{4} \\ a = \frac{3}{4} \end{matrix} \right] = \frac{1}{4a} \ln \left| \frac{t-a}{t+a} \right| + C = \frac{1}{3} \ln \left| \frac{x-\frac{1}{2}}{x+1} \right| + C = \frac{1}{3} \ln \left| \frac{2x-1}{x+1} \right| + C$