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Question-201184




Question Number 201184 by Calculusboy last updated on 01/Dec/23
Answered by Sutrisno last updated on 01/Dec/23
misal :  (√(2x))+4=u→dx=(√(2x))du  =∫((√(2x))/u).(√(2x))du  =∫(((u−4)^2 )/u)du  =∫((u^2 −8u+16)/u)du  =∫u−8+((16)/u)du  =(1/2)u^2 −8u+16lnu+c  =(1/2)((√(2x))+4)^2 −8((√(2x))+4)+16ln((√(2x))+4)+c
$${misal}\::\:\:\sqrt{\mathrm{2}{x}}+\mathrm{4}={u}\rightarrow{dx}=\sqrt{\mathrm{2}{x}}{du} \\ $$$$=\int\frac{\sqrt{\mathrm{2}{x}}}{{u}}.\sqrt{\mathrm{2}{x}}{du} \\ $$$$=\int\frac{\left({u}−\mathrm{4}\right)^{\mathrm{2}} }{{u}}{du} \\ $$$$=\int\frac{{u}^{\mathrm{2}} −\mathrm{8}{u}+\mathrm{16}}{{u}}{du} \\ $$$$=\int{u}−\mathrm{8}+\frac{\mathrm{16}}{{u}}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{u}^{\mathrm{2}} −\mathrm{8}{u}+\mathrm{16}{lnu}+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{2}{x}}+\mathrm{4}\right)^{\mathrm{2}} −\mathrm{8}\left(\sqrt{\mathrm{2}{x}}+\mathrm{4}\right)+\mathrm{16}{ln}\left(\sqrt{\mathrm{2}{x}}+\mathrm{4}\right)+{c} \\ $$
Commented by Calculusboy last updated on 01/Dec/23
nice solution sir
$$\boldsymbol{{nice}}\:\boldsymbol{{solution}}\:\boldsymbol{{sir}} \\ $$

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