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Question Number 43191 by MASANJA J last updated on 08/Sep/18
integrate by use a partial friction ∫((lnx)/((1+x)^2 ))
$${integrate}\:{by}\:{use}\:{a}\:{partial}\:{friction} \\ $$$$\int\frac{{lnx}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} } \\ $$
Commented by mondodotto@gmail.com last updated on 09/Sep/18
did you mean by parts?
$$\mathrm{did}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{by}\:\mathrm{parts}? \\ $$
Commented by maxmathsup by imad last updated on 08/Sep/18
let I = ∫ ((ln(x)dx)/((1+x)^2 )) by parts u^′ =(1/((1+x)^2 )) and v=ln(x) I =−((ln(x))/(1+x)) +∫ (1/((1+x)x))dx but ∫ (dx/(x(1+x))) =∫((1/x) −(1/(x+1)))dx=ln∣(x/(x+1))∣ +c ⇒ I =ln∣(x/(x+1))∣−((ln(x))/(1+x)) +c .
$${let}\:{I}\:=\:\int\:\:\frac{{ln}\left({x}\right){dx}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }\:\:{by}\:{parts}\:{u}^{'} =\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }\:{and}\:{v}={ln}\left({x}\right)\: \\ $$$${I}\:=−\frac{{ln}\left({x}\right)}{\mathrm{1}+{x}}\:+\int\:\:\:\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right){x}}{dx}\:{but}\:\int\:\:\:\frac{{dx}}{{x}\left(\mathrm{1}+{x}\right)}\:=\int\left(\frac{\mathrm{1}}{{x}}\:−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right){dx}={ln}\mid\frac{{x}}{{x}+\mathrm{1}}\mid\:+{c}\:\Rightarrow \\ $$$${I}\:={ln}\mid\frac{{x}}{{x}+\mathrm{1}}\mid−\frac{{ln}\left({x}\right)}{\mathrm{1}+{x}}\:+{c}\:. \\ $$
Commented by maxmathsup by imad last updated on 09/Sep/18
integration by parts .
$${integration}\:{by}\:{parts}\:. \\ $$
Answered by alex041103 last updated on 08/Sep/18
first we use IBP: ∫((lnx)/((1+x)^2 ))dx=∫−lnx d((1/(1+x)))= =−(((lnx)/(1+x)) −∫((d(lnx))/(1+x)))= =∫(dx/(x(1+x))) − ((ln(x))/(1+x)) (1/(x(1+x)))=(1/x)−(1/(1+x)) ⇒∫(dx/(x(1+x)))=∫(dx/x) − ∫((d(x+1))/(1+x))= =ln(x)−ln(x+1)=ln((x/(1+x))) ⇒∫((lnx)/((1+x)^2 ))dx=ln((x/(1+x)))−((ln(x))/(1+x))+C
$${first}\:{we}\:{use}\:{IBP}: \\ $$$$\int\frac{{lnx}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx}=\int−{lnx}\:{d}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)= \\ $$$$=−\left(\frac{{lnx}}{\mathrm{1}+{x}}\:−\int\frac{{d}\left({lnx}\right)}{\mathrm{1}+{x}}\right)= \\ $$$$=\int\frac{{dx}}{{x}\left(\mathrm{1}+{x}\right)}\:−\:\frac{{ln}\left({x}\right)}{\mathrm{1}+{x}} \\ $$$$\frac{\mathrm{1}}{{x}\left(\mathrm{1}+{x}\right)}=\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{1}+{x}} \\ $$$$\Rightarrow\int\frac{{dx}}{{x}\left(\mathrm{1}+{x}\right)}=\int\frac{{dx}}{{x}}\:−\:\int\frac{{d}\left({x}+\mathrm{1}\right)}{\mathrm{1}+{x}}= \\ $$$$={ln}\left({x}\right)−{ln}\left({x}+\mathrm{1}\right)={ln}\left(\frac{{x}}{\mathrm{1}+{x}}\right) \\ $$$$\Rightarrow\int\frac{{lnx}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx}={ln}\left(\frac{{x}}{\mathrm{1}+{x}}\right)−\frac{{ln}\left({x}\right)}{\mathrm{1}+{x}}+{C} \\ $$

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