integrate (x-1) dx/(x+1)[{(x^3)+(x^2)+x}^(1/2)]

Question

amartya mitra · Answer

multiply by (x+1) on the numerator as well as the denominator.then divide both the numerator and denominator by x^2. in the denominator we have(x+(1/x) +2)(x+(1/x)+1)^(1/2).we can substitute (x+1/x)to get(1-1/x^2)dx. we can substiute this in the numerator and then by subsequent minor substitutions , we can get the answer as (2 tan^(-1))(x+1/x +1)^1/2

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integrate (x-1) dx/(x+1)[{(x^3)+(x^2)+x}^(1/2)]

integrate (x-1) dx/(x+1)[{(x^3)+(x^2)+x}^(1/2)]

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