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Answer:$=\frac{1}{2}\left[\log \left|x^{2}+3 x+2\right|\right]-\frac{3}{2}\left[\log \left|\frac{x+1}{x+2}\right|\right]+c$

Hint Using Integration method

Given:$\int \frac{x}{x^{2}+3 x+2} d x$

Solution: Let,$x^{2}+3 x+2=t$

$(2 x+3) d x=d t$

\begin{aligned} &I=\frac{2}{2} \int \frac{x}{x^{2}+3 x+2} d x \\ &=\frac{1}{2} \int \frac{2 x}{x^{2}+3 x+2} d x \\ &=\frac{1}{2} \int \frac{2 x+3-3}{x^{2}+3 x+2} d x \\ &=\frac{1}{2}\left[\int \frac{2 x+3}{x^{2}+3 x+2} d x-\int \frac{3}{x^{2}+3 x+2} d x\right] \end{aligned}

$=\frac{1}{2}\left[\frac{d t}{t}-3 \int \frac{d x}{(x+1)(x+2)}\right]$                                    $\left[\begin{array}{c} \because\left(x^{2}+3 x+2\right) \\ =\left(x^{2}+2 x+x+2\right) \\ =(x(x+2)+1(x+2)) \\ =(x+1)(x+2) \end{array}\right]$

$=\frac{1}{2}\left[\log t-3 \int\left(\frac{1}{x+1}-\frac{1}{x+2}\right) d x\right]$                      $\left[\int \frac{1}{x} d x=\log x+c\right]$

$=\frac{1}{2}\left[\log \left(x^{2}+3 x+2\right)-3(\log (x+1)-\log (x+2)]+c\right.$

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

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

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