Question Number 5425 by love math last updated on 14/May/16
$$\int_{\mathrm{1}} ^{\mathrm{3}} {ln}\left(\mathrm{3}{x}−\mathrm{2}\right){dx} \\ $$
Answered by nchejane last updated on 14/May/16
![[xln(3x−2)]_1 ^3 −3∫_1 ^3 x(dx/(3x−2)) =[3ln(7)−3∫_1 ^3 [(1/3)+(2/3)(1/(3x−2)) ] dx =3ln(7)−[x+(2/3)ln(3x−2) ]_1 ^3 =3ln(7)−2−(2/3)ln(7) =(7/3)ln(7)−2](https://www.tinkutara.com/question/Q5429.png)
$$\left[{xln}\left(\mathrm{3}{x}−\mathrm{2}\right)\right]_{\mathrm{1}} ^{\mathrm{3}} −\mathrm{3}\int_{\mathrm{1}} ^{\mathrm{3}} {x}\frac{{dx}}{\mathrm{3}{x}−\mathrm{2}} \\ $$$$=\left[\mathrm{3}{ln}\left(\mathrm{7}\right)−\mathrm{3}\int_{\mathrm{1}} ^{\mathrm{3}} \left[\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}\frac{\mathrm{1}}{\mathrm{3}{x}−\mathrm{2}}\:\right]\:{dx}\right. \\ $$$$=\mathrm{3}{ln}\left(\mathrm{7}\right)−\left[{x}+\frac{\mathrm{2}}{\mathrm{3}}{ln}\left(\mathrm{3}{x}−\mathrm{2}\right)\:\right]_{\mathrm{1}} ^{\mathrm{3}} \\ $$$$=\mathrm{3}{ln}\left(\mathrm{7}\right)−\mathrm{2}−\frac{\mathrm{2}}{\mathrm{3}}{ln}\left(\mathrm{7}\right) \\ $$$$=\frac{\mathrm{7}}{\mathrm{3}}{ln}\left(\mathrm{7}\right)−\mathrm{2} \\ $$