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Question Number 25224 by kshreyasingh200@gmil.com last updated on 06/Dec/17

∫(2x−1)(√(x×x−x+1   dx))

$$\int\left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{{x}×{x}−{x}+\mathrm{1}\:\:\:{dx}} \\ $$

Answered by prakash jain last updated on 06/Dec/17

∫(2x−1)(√(x^2 −x+1)) dx  x^2 −x+1=u  (2x−1)dx=du  ∫(√u) du  =(2/3)u^(3/2) +C  =(2/3)(x^2 −x+1)^(3/2) +C

$$\int\left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:{dx} \\ $$$${x}^{\mathrm{2}} −{x}+\mathrm{1}={u} \\ $$$$\left(\mathrm{2}{x}−\mathrm{1}\right){dx}={du} \\ $$$$\int\sqrt{{u}}\:{du} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}{u}^{\mathrm{3}/\mathrm{2}} +{C} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} +{C} \\ $$

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