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Question: Evaluate \(\int \frac { d x } { x + \sqrt { x ^ { 2 } - x + 1 } }\) \(\left( \text{Where}t = \frac{\...

Evaluate ∫dxx+x2−x+1\int \frac { d x } { x + \sqrt { x ^ { 2 } - x + 1 } } (Wheret=x2−x+1+1x)\left( \text{Where}t = \frac{\sqrt{x^{2} - x + 1 + 1}}{x} \right)

A

2log⁡e∣t∣−12log⁡e∣t−1∣−32log⁡e∣t+1∣+3(t+1)+c2\log_{e}|t| - \frac{1}{2}\log_{e}|t - 1| - \frac{3}{2}\log_{e}|t + 1| + \frac{3}{(t + 1)} + c

B

2log⁡e∣t∣+12log⁡e∣t−1∣+32log⁡e∣t+1∣+3(t+1)+c2\log_{e}|t| + \frac{1}{2}\log_{e}|t - 1| + \frac{3}{2}\log_{e}|t + 1| + \frac{3}{(t + 1)} + c

C

2log⁡e∣t∣−12log⁡e∣t−1∣+32log⁡e∣t+1∣+3(t+1)+c2\log_{e}|t| - \frac{1}{2}\log_{e}|t - 1| + \frac{3}{2}\log_{e}|t + 1| + \frac{3}{(t + 1)} + c

D

None of these

Answer

2log⁡e∣t∣−12log⁡e∣t−1∣−32log⁡e∣t+1∣+3(t+1)+c2\log_{e}|t| - \frac{1}{2}\log_{e}|t - 1| - \frac{3}{2}\log_{e}|t + 1| + \frac{3}{(t + 1)} + c

Explanation

Solution

Since here c=1,c = 1, we can apply the second Euler substitution.

x2−x+1=tx−1\sqrt{x^{2} - x + 1} = tx - 1

Hence (2t−1)x=(t2−1)x2(2t - 1)x = (t^{2} - 1)x^{2}; x=2t−1t2−1x = \frac{2t - 1}{t^{2} - 1}

∴dx=2(t2−t+1)dt(t2−1)2\therefore dx = \frac{2(t^{2} - t + 1)dt}{(t^{2} - 1)^{2}} and x+x2−x+1=1t−1x + \sqrt{x^{2} - x + 1} = \frac{1}{t - 1}

∴I=∫dxx+x2−x+1⥂=∫−2t2+2t−2t(t−1)(t+1)2⥂dt⥂\therefore I = \int_{}^{}\frac{dx}{x + \sqrt{x^{2} - x + 1}} ⥂ = \int_{}^{}{\frac{- 2t^{2} + 2t - 2}{t(t - 1)(t + 1)^{2}} ⥂ dt} ⥂Using partial

fractions, we have,

−2t2+2t−2t(t−1)(t+1)2=At+Bt−1+C(t+1)+D(t+1)2\frac{- 2t^{2} + 2t - 2}{t(t - 1)(t + 1)^{2}} = \frac{A}{t} + \frac{B}{t - 1} + \frac{C}{(t + 1)} + \frac{D}{(t + 1)^{2}}

or (−2t+2t−2)=A(t−1)(t+1)2+B(t+1)2+C(t−1)(t+1)t+Dt.( - 2t + 2t - 2) = A(t - 1)(t + 1)^{2} + B(t + 1)^{2} + C(t - 1)(t + 1)t + Dt. We get A=2,B=1/2,C=−3/2,D=−3.A = 2,B = 1/2,C = - 3/2,D = - 3.

Hence (Wheret=x2−x+1+1x)\left( \text{Where}t = \frac{\sqrt{x^{2} - x + 1 + 1}}{x} \right)