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 93481 by mashallah last updated on 13/May/20

$$\int \left(\log \mathrm{x}/{\mathrm{x}}^2\right)\mathrm{dx}=$$

Commented by abdomathmax last updated on 15/May/20

$$I=\int \frac{lnx}{x^2}dxbyparts$$$$I=-\frac{lnx}{x}-\int (-\frac{1}{x})\times \frac{dx}{x}=-\frac{lnx}{x}+\int \frac{dx}{x^2}$$$$=-\frac{lnx}{x}-\frac{1}{x}+C=-\frac{1+lnx}{x}+C$$

Answered by john santu last updated on 13/May/20

$$\int \frac{\ln \mathrm{x}}{{\mathrm{x}}^2}\mathrm{dx}=\int \ln \left(\mathrm{x}\right).{\mathrm{x}}^{-2}\mathrm{dx}=\mathrm{W}$$$$\left[\mathrm{by}\mathrm{parts}\right]$$$$\mathrm{u}=\ln \left(\mathrm{x}\right)\Rightarrow \mathrm{du}=\frac{\mathrm{dx}}{\mathrm{x}}$$$$\mathrm{v}=\int {\mathrm{x}}^{-2}\mathrm{dx}=-{\mathrm{x}}^{-1}$$$$\mathrm{W}=-{\mathrm{x}}^{-1}\ln \left(\mathrm{x}\right)+\int {\mathrm{x}}^{-1}.\frac{\mathrm{dx}}{\mathrm{x}}$$$$\mathrm{W}=-\frac{\ln \left(\mathrm{x}\right)}{\mathrm{x}}-\frac{1}{\mathrm{x}}+\mathrm{c}$$$$\mathrm{W}=-\frac{\ln \left(\mathrm{x}\right)+1}{\mathrm{x}}+\mathrm{c}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com