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x-1-2-ln-1-1-x-x-dx-




Question Number 135646 by metamorfose last updated on 14/Mar/21
∫(x+(1/2))ln(1+(1/x))−x dx=...?
(x+12)ln(1+1x)xdx=?
Answered by Ñï= last updated on 15/Mar/21
∫[(x+(1/2))ln(1+(1/x))−x]dx  =∫[(x+(1/2))(ln(x+1)−lnx)−x]dx  =∫(x+(1/2))ln(x+1)dx−∫(x+(1/2))lnxdx−∫xdx  =((1/2)x^2 +(1/2)x)ln(x+1)−∫((x+1−(1/2))/(x+1))dx−((1/2)x^2 +(1/2)x)lnx+∫((x+(1/2))/x)dx−(1/2)x^2   =((1/2)x^2 +(1/2)x)ln(1+(1/x))+(1/2)ln(x^2 +x)−(1/2)x^2 +C
[(x+12)ln(1+1x)x]dx=[(x+12)(ln(x+1)lnx)x]dx=(x+12)ln(x+1)dx(x+12)lnxdxxdx=(12x2+12x)ln(x+1)x+112x+1dx(12x2+12x)lnx+x+12xdx12x2=(12x2+12x)ln(1+1x)+12ln(x2+x)12x2+C

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