Question

Solve $\int \frac{5x-1}{3x^2+x+2}dx$

Answer 1

Consider

$\int \frac{5x-1}{3x^2+x+2}dx$

$=\int \frac{6x+1-x-2}{3x^2+x+2}dx$

$=\int \frac{6x+1}{3x^2+x+2}dx-\frac{1}{6}\int \frac{6x+1}{3x^2+x+2}dx-\frac{1}{6}\int \frac{11}{3x^2+x+2}dx$

$=\frac{5}{6}\int \frac{6x+1}{3x^2+x+2}dx-\frac{11}{6}\int \frac{dx}{3x^2+x+2}$

$=\frac{5}{6}\log (3x^2+x+2)-\frac{11}{6}\int \frac{dx}{3(x^2+\frac{x}{3}+\frac{2}{3})}$

$=\frac{5}{6}\log (3x^2+x+2)-\frac{11}{18}\int \frac{dx}{(x^2+2\frac{x}{6}+\frac{1}{36}-\frac{1}{36}+\frac{2}{3})}$

$=\frac{5}{6}\log (3x^2+x+2)-\frac{11}{18}\int \frac{dx}{({(x+\frac{1}{6})}^2+\frac{23}{36})}$

$=\frac{5}{6}\log (3x^2+x+2)-\frac{11}{18}\times \frac{6}{\sqrt{23}}{\tan }^{-1}\left(\frac{x+\frac{1}{6}}{\frac{\sqrt{23}}{6}}\right)+C$

$=\frac{5}{6}\log (3x^2+x+2)-\frac{11}{3\sqrt{23}}{\tan }^{-1}\left(\frac{6x+1}{\sqrt{23}}\right)+C$