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Question Number 42088 by maxmathsup by imad last updated on 17/Aug/18
find ∫ (1+(1/x^2 ))ln(1−(1/x))dx .
find∫(1+1x2)ln(1−1x)dx.
Commented by maxmathsup by imad last updated on 17/Aug/18
let A = ∫ (1+(1/x^2 ))ln(1−(1/x))dx by parts u^′ =1+(1/x^2 ) and v=ln(1−(1/x)) A = (x−(1/x))ln(1−(1/x)) −∫ (x−(1/x)) (1/(x^2 (1−(1/x))))dx =(x−(1/x))ln(1−(1/x)) − ∫ ((x^2 −1)/x) (1/(x^2 −x)) dx =(x−(1/x))ln(1−(1/x)) −∫ (((x−1)(x+1))/(x^2 (x−1))) dx =(x−(1/x))ln(1−(1/x)) −∫ ((x+1)/x^2 ) dx A=(x−(1/x))ln(1−(1/x))−ln∣x∣ +(1/x) +c
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+(1/x^2 ))ln(1−(1/x))dx t=1−(1/x) dt=(1/x^2 )dx ∫ln(1−(1/x))dx+∫(1/x^2 )ln(1−(1/x))dx I_1 +I_2 I_2 =∫(1/x^2 )ln(1−(1/x))dx =∫lnt dt tlnt−∫(1/t)tdt tlnt−t+c_2 (1−(1/x))ln(1−(1/x))−(1−(1/x))+c_2 I_1 =∫ln(1−(1/x))dx =xln(1−(1/x))−∫((0+(1/x^2 ))/(1−(1/x)))xdx =xln(1−(1/x))−∫((xdx)/(x^2 (((x−1)/x)))) =xln(1−(1/x))−∫(dx/(x−1)) =xln(1−(1/x))−ln(x−1)+c_1 =2xln(1−(1/x))−ln(x−1)+(1−(1/x))+c is ans
∫(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

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