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 117841 by john santu last updated on 14/Oct/20

∫ ln (1−e^(−2x) ) dx =?

∫ln(1−e−2x)dx=?

Answered by MJS_new last updated on 14/Oct/20

∫ln (1−e^(−2x) ) dx= [t=e^(−2x) → dx=−(e^(2x) /2)dt] =−(1/2)∫((ln (1−t))/t)dt=(1/2)Li_2 t =(1/2)Li_2 e^(−2x) +C

∫ln(1−e−2x)dx=[t=e−2x→dx=−e2x2dt]=−12∫ln(1−t)tdt=12Li2t=12Li2e−2x+C

Commented by john santu last updated on 14/Oct/20

thank prof

thankprof

Terms of Service

Privacy Policy

Contact: info@tinkutara.com