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Question Number 167233 by mkam last updated on 10/Mar/22

∫ ((lnx)/(x+1))dx

$$\int\:\frac{{lnx}}{{x}+\mathrm{1}}{dx} \\ $$

Answered by mindispower last updated on 10/Mar/22

=ln(x)ln(1+x)−∫((ln(1+x))/x)dx  =ln(x)ln(1+x)−∫((ln(1−(−x)))/(−x))d(−x)  =ln(x)ln(1+x)+Li_2 (−x)+c

$$={ln}\left({x}\right){ln}\left(\mathrm{1}+{x}\right)−\int\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx} \\ $$$$={ln}\left({x}\right){ln}\left(\mathrm{1}+{x}\right)−\int\frac{{ln}\left(\mathrm{1}−\left(−{x}\right)\right)}{−{x}}{d}\left(−{x}\right) \\ $$$$={ln}\left({x}\right){ln}\left(\mathrm{1}+{x}\right)+{Li}_{\mathrm{2}} \left(−{x}\right)+{c} \\ $$

Answered by ArnabIndia last updated on 10/Mar/22

Commented by ArnabIndia last updated on 10/Mar/22

i dont know why the picture is showing blurr

Commented by mkam last updated on 11/Mar/22

are you have page on face book app

$${are}\:{you}\:{have}\:{page}\:{on}\:{face}\:{book}\:{app} \\ $$

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