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Question Number 43191 by MASANJA J last updated on 08/Sep/18

$$integratebyuseapartialfriction$$$$\int \frac{lnx}{{(1+x)}^2}$$

Commented by mondodotto@gmail.com last updated on 09/Sep/18

$$\mathrm{did}\mathrm{you}\mathrm{mean}\mathrm{by}\mathrm{parts}?$$

Commented by maxmathsup by imad last updated on 08/Sep/18

$$letI=\int \frac{ln\left(x\right)dx}{{(1+x)}^2}byparts{u}^{{}^{\prime }}=\frac{1}{{(1+x)}^2}andv=ln\left(x\right)$$$$I=-\frac{ln\left(x\right)}{1+x}+\int \frac{1}{(1+x)x}dxbut\int \frac{dx}{x(1+x)}=\int (\frac{1}{x}-\frac{1}{x+1})dx=ln\mid \frac{x}{x+1}\mid +c\Rightarrow$$$$I=ln\mid \frac{x}{x+1}\mid -\frac{ln\left(x\right)}{1+x}+c.$$

Commented by maxmathsup by imad last updated on 09/Sep/18

$$integrationbyparts.$$

Answered by alex041103 last updated on 08/Sep/18

$$firstweuseIBP:$$$$\int \frac{lnx}{{(1+x)}^2}dx=\int -lnxd\left(\frac{1}{1+x}\right)=$$$$=-(\frac{lnx}{1+x}-\int \frac{d\left(lnx\right)}{1+x})=$$$$=\int \frac{dx}{x(1+x)}-\frac{ln\left(x\right)}{1+x}$$$$\frac{1}{x(1+x)}=\frac{1}{x}-\frac{1}{1+x}$$$$\Rightarrow \int \frac{dx}{x(1+x)}=\int \frac{dx}{x}-\int \frac{d(x+1)}{1+x}=$$$$=ln\left(x\right)-ln(x+1)=ln\left(\frac{x}{1+x}\right)$$$$\Rightarrow \int \frac{lnx}{{(1+x)}^2}dx=ln\left(\frac{x}{1+x}\right)-\frac{ln\left(x\right)}{1+x}+C$$

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