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∫(1+1x2)ln(1−1x)dx.
Commented by maxmathsup by imad last updated on 17/Aug/18
letA=∫(1+1x2)ln(1−1x)dxbypartsu′=1+1x2andv=ln(1−1x)A=(x−1x)ln(1−1x)−∫(x−1x)1x2(1−1x)dx=(x−1x)ln(1−1x)−∫x2−1x1x2−xdx=(x−1x)ln(1−1x)−∫(x−1)(x+1)x2(x−1)dx=(x−1x)ln(1−1x)−∫x+1x2dxA=(x−1x)ln(1−1x)−ln∣x∣+1x+c
Answered by tanmay.chaudhury50@gmail.com last updated on 17/Aug/18
∫(1+1x2)ln(1−1x)dxt=1−1xdt=1x2dx∫ln(1−1x)dx+∫1x2ln(1−1x)dxI1+I2I2=∫1x2ln(1−1x)dx=∫lntdttlnt−∫1ttdttlnt−t+c2(1−1x)ln(1−1x)−(1−1x)+c2I1=∫ln(1−1x)dx=xln(1−1x)−∫0+1x21−1xxdx=xln(1−1x)−∫xdxx2(x−1x)=xln(1−1x)−∫dxx−1=xln(1−1x)−ln(x−1)+c1=2xln(1−1x)−ln(x−1)+(1−1x)+cisans
Terms of Service
Privacy Policy
Contact: info@tinkutara.com