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


Let $5 x-2=A \frac{d}{d x}\left(1+2 x+3 x^{2}\right)+B$

$\Rightarrow 5 x-2=A(2+6 x)+B$

Equating the coefficient of x and constant term on both sides, we obtain

$5=6 A \Rightarrow A=\frac{5}{6}$

$2 A+B=-2 \Rightarrow B=-\frac{11}{3}$


$\therefore 5 x-2=\frac{5}{6}(2+6 x)+\left(-\frac{11}{3}\right)$

$\Rightarrow \int \frac{5 x-2}{1+2 x+3 x^{2}} d x=\int \frac{\frac{5}{6}(2+6 x)-\frac{11}{3}}{1+2 x+3 x^{2}} d x$

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

Let $I_{1}=\int \frac{2+6 x}{1+2 x+3 x^{2}} d x$ and $I_{2}=\int \frac{1}{1+2 x+3 x^{2}} d x$

$\therefore \int \frac{5 x-2}{1+2 x+3 x^{2}} d x=\frac{5}{6} I_{1}-\frac{11}{3} I_{2}$   ....(1)

$I_{1}=\int \frac{2+6 x}{1+2 x+3 x^{2}} d x$

Let $1+2 x+3 x^{2}=t$

$\Rightarrow(2+6 x) d x=d t$

$\therefore I_{1}=\int \frac{d t}{t}$

$I_{1}=\log |t|$

$I_{1}=\log \left|1+2 x+3 x^{2}\right|$    ...(2)

$I_{2}=\int \frac{1}{1+2 x+3 x^{2}} d x$

$1+2 x+3 x^{2}$ can be written as $1+3\left(x^{2}+\frac{2}{3} x\right)$


$1+3\left(x^{2}+\frac{2}{3} x\right)$

$=1+3\left(x^{2}+\frac{2}{3} x+\frac{1}{9}-\frac{1}{9}\right)$





$I_{2}=\frac{1}{3} \int \frac{1}{\left[\left(x+\frac{1}{3}\right)^{2}+\left(\frac{\sqrt{2}}{3}\right)^{2}\right]^{d x}}$

$=\frac{1}{3}\left[\frac{1}{\frac{\sqrt{2}}{3}} \tan ^{-1}\left(\frac{x+\frac{1}{3}}{\frac{\sqrt{2}}{2}}\right)\right]$

$=\frac{1}{3}\left[\frac{3}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)\right]$

$=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)$      ...(3)

Substituting equations (2) and (3) in equation (1), we obtain

$\int \frac{5 x-2}{1+2 x+3 x^{2}} d x=\frac{5}{6}\left[\log \left|1+2 x+3 x^{2}\right|\right]-\frac{11}{3}\left[\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)\right]+\mathrm{C}$

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


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