$=\int\limits_0^1 {x\,\,\ln (x + 2)dx\,} \,\, - \,\,\ln 2\,\,\int\limits_0^1 {x\,\,dx\,} \,$
$ \Rightarrow$ $\ln (x + 2)\,.\,\left. {\frac{{{x^2}}}{2}} \right]_0^1\,\, - \,\,\int\limits_0^1 {\frac{{{x^2}}}{{x + 2}}\,\,\,dx\, - \,\,\frac{{\ln 2}}{2}} \,$
$=\frac{1}{2}\,\ln \,3\,\,\, - \,\,\,\int\limits_0^1 {\frac{{{x^2} - 4 + 4}}{{x + 2}}\,\,\,dx\, - \,\,\frac{{\ln 2}}{2}} \,$
$\Rightarrow$ $\frac{1}{2}\,\ln \,\frac{3}{2}\,\,\, - \,\,\,\int\limits_0^1 {\left( {(x - 2)\,\,\, + \,\,\frac{4}{{x + 2}}} \right)\,\,\,dx\,} $
now proceed
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The solution of the equation $(2\text{y}-1)\text{dx}-(2\text{x}+3)\text{dy}=0$ is:
$\frac{2\text{x}-1}{2\text{y}+3}=\text{k}$
$\frac{2\text{y}+1}{2\text{x}-3}=\text{k}$
$\frac{2\text{x}+3}{2\text{y}-1}=\text{k}$
$\frac{2\text{x}-1}{2\text{y}-1}=\text{k}$