Let $I=\int \frac{5x-2}{1+2x+3x^2}dx$ Put $5x-2 = A\left(\frac{d}{dx}(3x^2+2x+1)\right)+B$ $5x - 2 = A(6x+2) + B$ ... (i) Comparing coefficients of x in (i), we get $5 = 6A \Rightarrow A = \frac{5}{6}$ Comparing the constant terms in (i), we get $-2 = 2A + B$ $\Rightarrow -2 = 2 \times \frac{5}{6} + B$ $\Rightarrow -2 = \frac{5}{3} + B$ $\Rightarrow B - \frac{11}{3}$ $\therefore I = \int \frac{\frac{5}{6}(6x+2) - \frac{11}{3}}{3x^2+2x+1} dx$ $= \frac{5}{6} \int \frac{6x+2}{3x^2+2x+1} dx - \frac{11}{3} \int \frac{dx}{3x^2+2x+1}$ Let $I = \frac{5}{6} I_1 - \frac{11}{3} I_2$ ... (ii) where, $I_1 = \int \frac{6x+2}{3x^2+2x+1} dx$ Put $3x^2 + 2x + 1 = t$ $\Rightarrow (6x+2)dx = dt$ $\therefore I_1 = \int \frac{dt}{t} = \log\lvert t\rvert + C_1$ $= \log\lvert 3x^2 + 2x + 1\rvert + C_1$ ... (iii) and $I_2 = \int \frac{dx}{3x^2+2x+1}$ $= \frac{1}{3} \int \frac{dx}{x^2 + \frac{2}{3}x + \frac{1}{3}}$$= \frac{1}{3} \int \frac{dx}{x^2 + \frac{2}{3}x + \frac{1}{9} - \frac{1}{9} + \frac{1}{3}}$$= \frac{1}{3} \int \frac{dx}{\left(x+\frac{1}{3}\right)^2 + \frac{2}{9}}$$= \frac{1}{3} \int \frac{dx}{\left(x+\frac{1}{3}\right)^2 + \left(\frac{\sqrt{2}}{9}\right)^2}$$= \frac{1}{3} \times \frac{1}{\frac{\sqrt{2}}{3}} \tan^{-1}\left(\frac{x+\frac{1}{3}}{\frac{\sqrt{2}}{3}}\right)$$= \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{3x+1}{\sqrt{2}}\right) + C_2$... (iv)From (ii), (iii) and (iv), we get