$\int \frac{5\mathrm{x}-2}{1+2\mathrm{x}+3{\mathrm{x}}^2}\mathrm{dx}$ is equal to 

$\int \frac{5\mathrm{x}-2}{1+2\mathrm{x}+3{\mathrm{x}}^2}\mathrm{dx}$
Let $5\mathrm{x}-2=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}\left(1+2\mathrm{x}+3{\mathrm{x}}^2\right)+\mathrm{B}$
$\Rightarrow 5\mathrm{x}-2=\mathrm{A}(2+6\mathrm{x})+\mathrm{B}\Rightarrow 5\mathrm{x}-2=6\mathrm{Ax}+(2\mathrm{A}+\mathrm{B})$
On equating the coefficient of x and constant on both sides, we get
$5=6\mathrm{A}\Rightarrow \mathrm{A}=\frac{5}{6}$and $2\mathrm{A}+\mathrm{B}=-2\Rightarrow \mathrm{B}=\frac{-11}{3}$
$\begin{array}{r}\therefore \mathrm{l}=\int \frac{5\mathrm{x}-2}{1+2\mathrm{x}+3{\mathrm{x}}^2}\mathrm{dx}=\int \frac{\frac{3}{6}(2+6\mathrm{x})-\frac{11}{3}}{1+2\mathrm{x}+3{\mathrm{x}}^2}\mathrm{dx}\\ =\frac{5}{6}\int \frac{2+6\mathrm{x}}{1+2\mathrm{x}+3{\mathrm{x}}^2}\mathrm{dx}-\frac{11}{3}\int \frac{\mathrm{dx}}{1+2\mathrm{x}+3{\mathrm{x}}^2}\end{array}$
Let 
$\begin{array}{l}{\mathrm{I}}_1=\int \frac{2+6\mathrm{x}}{1+2\mathrm{x}+3{\mathrm{x}}^2}\text{ and }{\mathrm{I}}_2=\int \frac{\mathrm{dx}}{1+2\mathrm{x}+3{\mathrm{x}}^2}\\ \therefore \mathrm{I}=\frac{5}{6}{\mathrm{l}}_1-\frac{11}{3}{\mathrm{I}}_2 ....\left(\mathrm{i}\right) \end{array}$
Now, ${\mathrm{I}}_1=\int \frac{2+6\mathrm{x}}{1+2\mathrm{x}+3{\mathrm{x}}^2}\mathrm{dx}$
Let $1+2\mathrm{x}+3{\mathrm{x}}^2=\mathrm{t}\Rightarrow (2+6\mathrm{x})\mathrm{dx}=\mathrm{dt}$
$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{I}}_1=\int \frac{\mathrm{dt}}{\mathrm{t}}=\log \left|\mathrm{t}\right|+{\mathrm{C}}_1\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{I}}_1=\log \left|1+2\mathrm{x}+3{\mathrm{x}}^2\right|+{\mathrm{C}}_1 ....\left(\mathrm{ii}\right)\end{array}$
Also, ${\mathrm{I}}_2=\int \frac{\mathrm{dx}}{1+2\mathrm{x}+3{\mathrm{x}}^2}$
$1+2\mathrm{x}+3{\mathrm{x}}^2 \mathrm{can} \mathrm{be} \mathrm{written} \mathrm{as} 1+3\left({\mathrm{x}}^2+\frac{2}{3}\mathrm{x}\right).$
Therefore,
$\begin{array}{l}1+3\left({\mathrm{x}}^2+\frac{2}{3}\mathrm{x}\right)=1+3\left({\mathrm{x}}^2+\frac{2}{3}\mathrm{x}+\frac{1}{9}-\frac{1}{9}\right)\\ =1+3{\left(\mathrm{x}+\frac{1}{3}\right)}^2-\frac{1}{3}=\frac{2}{3}+3{\left(\mathrm{x}+\frac{1}{3}\right)}^2\\ =3\left[{\left(\mathrm{x}+\frac{1}{3}\right)}^2+\frac{2}{9}\right]=3\left[{\left(\mathrm{x}+\frac{1}{3}\right)}^2+{\left(\frac{\sqrt{2}}{3}\right)}^2\right]\end{array}$
$\begin{array}{r}\left.\therefore {\mathrm{I}}_2=\frac{1}{3}\int \frac{\mathrm{dx}}{\left[{\left(\mathrm{x}+\frac{1}{3}\right)}^2+{\left(\frac{\sqrt{2}}{3}\right)}^2\right]}=\frac{1}{3}\mid \frac{1}{\sqrt{2}/3}{\tan }^{-1}\left(\frac{\mathrm{x}+\frac{1}{3}}{\sqrt{2}/3}\right)\right]+{\mathrm{C}}_2\\ \left[\because \int \frac{\mathrm{dx}}{{\mathrm{a}}^2+{\mathrm{x}}^2}=\frac{1}{\mathrm{a}}{\tan }^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\right]\end{array}$
$\begin{array}{l}=\frac{1}{3}\left[\frac{3}{\sqrt{2}}{\tan }^{-1}\left(\frac{3\mathrm{x}+1}{\sqrt{2}}\right)\right]+{\mathrm{C}}_2\\ =\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{3\mathrm{x}+1}{\sqrt{2}}\right)+{\mathrm{C}}_{2 } ....\left(\mathrm{iii}\right) \end{array}$
On substituting the values of l₁ and l₂ from Eq. (ii) and Eq. (i),we get
$\begin{array}{l}\mathrm{I}=\frac{5}{6}\left[\log \left|1+2\mathrm{x}+3{\mathrm{x}}^2\right|\right]-\frac{11}{3}\left[\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{3\mathrm{x}+1}{\sqrt{2}}\right)\right]+\mathrm{C}\\ \left[\because \frac{5}{6}{\mathrm{C}}_1-\frac{11}{3}{\mathrm{C}}_2=\mathrm{C}\right]\\ =\frac{5}{6}\log \left|1+2\mathrm{x}+3{\mathrm{x}}^2\right|-\frac{11}{3\sqrt{2}}{\tan }^{-1}\left(\frac{3\mathrm{x}+1}{\sqrt{2}}\right)+\mathrm{C}\end{array}$