Evaluate ∫dxx+x2−x+1

Since here c = 1, we can apply the second Eulersubstitution x2−x+1=tx−1Hence (2t−1)x=t2−1x2;x=2t−1t2−1∴ dx=2t2−t+1dtt2−12 and x+x2−x+1=1t−1∴ I=∫dxx+x2−x+1=∫−2t2+2t−2t(t−1)(t+1)2dtUsing partial fractions, we have−2t2+2t−2t(t−1)(t+1)2=At+Bt−1+C(t+1)+D(t+1)2 or (−2t+2t−2)=A(t−1)(t+1)2+B(t+1)2+C(t−1)(t+1)t+DtWe get A=2,B=1/2,C=−3/2,D=−3.Hence I=2∫dtt−12∫dtt−1−32∫dt(t+1)−3∫dt(t+1)2=2loge⁡|t|−12loge⁡|t−1|−32loge⁡|t+1|+3(t+1)+c Where t=x2-x+1+1x

Evaluate ∫dxx+x2−x+1

Since here c = 1, we can apply the second Eulersubstitution x2−x+1=tx−1Hence (2t−1)x=t2−1x2;x=2t−1t2−1∴ dx=2t2−t+1dtt2−12 and x+x2−x+1=1t−1∴ I=∫dxx+x2−x+1=∫−2t2+2t−2t(t−1)(t+1)2dtUsing partial fractions, we have−2t2+2t−2t(t−1)(t+1)2=At+Bt−1+C(t+1)+D(t+1)2 or (−2t+2t−2)=A(t−1)(t+1)2+B(t+1)2+C(t−1)(t+1)t+DtWe get A=2,B=1/2,C=−3/2,D=−3.Hence I=2∫dtt−12∫dtt−1−32∫dt(t+1)−3∫dt(t+1)2=2loge⁡|t|−12loge⁡|t−1|−32loge⁡|t+1|+3(t+1)+c Where t=x2-x+1+1x

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Evaluate $\int \frac{d x}{x + \sqrt{x^{2} - x + 1}}$

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$2 \log_{e} | t | - \frac{1}{2} \log_{e} | t - 1 | - \frac{3}{2} \log_{e} | t + 1 | + \frac{3}{(t + 1)} + c$

$2 \log_{e} | t | + \frac{1}{2} \log_{e} | t - 1 | + \frac{3}{2} \log_{e} | t + 1 | + \frac{3}{(t + 1)} + c$

$2 \log_{e} | t | - \frac{1}{2} \log_{e} | t - 1 | + \frac{3}{2} \log_{e} | t + 1 | + \frac{3}{(t + 1)} + c$

None of these $(Where t = \frac{\sqrt{x^{2} - x + 1} + 1}{x})$

answer is A.

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Detailed Solution

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

substitution $\sqrt{x^{2} - x + 1} = t x - 1$

Hence $(2 t - 1) x = (t^{2} - 1) x^{2}; x = \frac{2 t - 1}{t^{2} - 1}$

$\begin{array}{l} ∴ d x = \frac{2 (t^{2} - t + 1) d t}{{(t^{2} - 1)}^{2}} and x + \sqrt{x^{2} - x + 1} = \frac{1}{t - 1} \\ ∴ I = \int \frac{d x}{x + \sqrt{x^{2} - x + 1}} = \int \frac{- 2 t^{2} + 2 t - 2}{t (t - 1) (t + 1)^{2}} d t \end{array}$

Using partial fractions, we have

$\begin{array}{l} \frac{- 2 t^{2} + 2 t - 2}{t (t - 1) (t + 1)^{2}} = \frac{A}{t} + \frac{B}{t - 1} + \frac{C}{(t + 1)} + \frac{D}{(t + 1)^{2}} or (- 2 t + 2 t - 2) \\ = A (t - 1) (t + 1)^{2} + B (t + 1)^{2} + C (t - 1) (t + 1) t + D t \end{array}$

We get $A = 2, B = 1 / 2, C = - 3 / 2, D = - 3$ .

Hence $I = 2 \int \frac{d t}{t} - \frac{1}{2} \int \frac{d t}{t - 1} - \frac{3}{2} \int \frac{d t}{(t + 1)} - 3 \int \frac{d t}{(t + 1)^{2}}$

$\begin{array}{l} = 2 \log_{e} | t | - \frac{1}{2} \log_{e} | t - 1 | - \frac{3}{2} \log_{e} | t + 1 | + \frac{3}{(t + 1)} + c \\ (Where t = \frac{\sqrt{x^{2} - x + 1} + 1}{x}) \end{array}$