Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 42088 by maxmathsup by imad last updated on 17/Aug/18

$$find\int (1+\frac{1}{x^2})ln(1-\frac{1}{x})dx.$$

Commented by maxmathsup by imad last updated on 17/Aug/18

$$letA=\int (1+\frac{1}{x^2})ln(1-\frac{1}{x})dxbyparts{u}^{{}^{\prime }}=1+\frac{1}{x^2}andv=ln(1-\frac{1}{x})$$$$A=(x-\frac{1}{x})ln(1-\frac{1}{x})-\int (x-\frac{1}{x})\frac{1}{x^2(1-\frac{1}{x})}dx$$$$=(x-\frac{1}{x})ln(1-\frac{1}{x})-\int \frac{x^2-1}{x}\frac{1}{x^2-x}dx$$$$=(x-\frac{1}{x})ln(1-\frac{1}{x})-\int \frac{(x-1)(x+1)}{x^2(x-1)}dx$$$$=(x-\frac{1}{x})ln(1-\frac{1}{x})-\int \frac{x+1}{x^2}dx$$$$A=(x-\frac{1}{x})ln(1-\frac{1}{x})-ln\mid x\mid +\frac{1}{x}+c$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Aug/18

$$\int (1+\frac{1}{x^2})ln(1-\frac{1}{x})dx$$$$t=1-\frac{1}{x}dt=\frac{1}{x^2}dx$$$$\int ln(1-\frac{1}{x})dx+\int \frac{1}{x^2}ln(1-\frac{1}{x})dx$$$$I_1+I_2$$$$I_2=\int \frac{1}{x^2}ln(1-\frac{1}{x})dx$$$$=\int lntdt$$$$tlnt-\int \frac{1}{t}tdt$$$$tlnt-t+c_2$$$$(1-\frac{1}{x})ln(1-\frac{1}{x})-(1-\frac{1}{x})+c_2$$$$I_1=\int ln(1-\frac{1}{x})dx$$$$=xln(1-\frac{1}{x})-\int \frac{0+\frac{1}{x^2}}{1-\frac{1}{x}}xdx$$$$$$$$=xln(1-\frac{1}{x})-\int \frac{xdx}{x^2\left(\frac{x-1}{x}\right)}$$$$=xln(1-\frac{1}{x})-\int \frac{dx}{x-1}$$$$=xln(1-\frac{1}{x})-ln(x-1)+c_1$$$$=2xln(1-\frac{1}{x})-ln(x-1)+(1-\frac{1}{x})+cisans$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com