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If-x-log-1-1-x-dx-f-x-log-x-1-g-x-x-2-Ax-C-then-




Question Number 55779 by gunawan last updated on 04/Mar/19
If ∫ x log (1+(1/x))dx         = f(x) ∙ log (x+1)+g(x) ∙ x^2 +Ax+C,  then
$$\mathrm{If}\:\int\:{x}\:\mathrm{log}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right){dx}\: \\ $$$$\:\:\:\:\:\:=\:{f}\left({x}\right)\:\centerdot\:\mathrm{log}\:\left({x}+\mathrm{1}\right)+{g}\left({x}\right)\:\centerdot\:{x}^{\mathrm{2}} +{Ax}+{C}, \\ $$$$\mathrm{then} \\ $$
Commented by gunawan last updated on 04/Mar/19
Find A
$$\mathrm{Find}\:{A} \\ $$
Commented by gunawan last updated on 04/Mar/19
A=  f(x)=  g(x)=
$${A}= \\ $$$${f}\left({x}\right)= \\ $$$${g}\left({x}\right)= \\ $$
Answered by MJS last updated on 04/Mar/19
∫xlog (1+(1/x)) dx=  =−(1/2)log (x+1) +x^2 ln (1+(1/x)) +(x/2)+C  (integration by parts)  f(x)=−(1/2)  g(x)=ln (1+(1/x))  A=(1/2)
$$\int{x}\mathrm{log}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\:{dx}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\left({x}+\mathrm{1}\right)\:+{x}^{\mathrm{2}} \mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\:+\frac{{x}}{\mathrm{2}}+{C} \\ $$$$\left(\mathrm{integration}\:\mathrm{by}\:\mathrm{parts}\right) \\ $$$${f}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${g}\left({x}\right)=\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right) \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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