Let x+1=Addxx2+x+2+Bx+1=A(2x+1)+Bx+1=2Ax+A+B On comparing 2A=1 A+B=1A=12 B=12∴∫(x+1)x2+x+2dx=∫12(x+1)+12x2+x+2dx=12∫(2x+1)x2+x+2dx+12x2+x+2dx=∴∫f′(x)f(x)dx=23f32(x)+c=12⋅23x2+x+232+12∫x2+2x⋅12+14−14+2dx=13x2+x+232+12∫x+122+722dx=13x2+x+232+12x+122x+122+722+7222sin⁡x−1x+1272+c=13x2+x+232+12(2x+1)4x+122+74+78sin⁡x−12x+17+c

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$\int (x - 1) \sqrt{x^{2} + x + 2} d x =$

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$\frac{1}{3} {(x^{2} + x + 2)}^{\frac{3}{2}} + \frac{1}{2} \cos^{- 1} (\frac{x + \frac{1}{2}}{\frac{\sqrt{7}}{2}}) + c$

$\frac{1}{3} {(x^{2} + x + 2)}^{\frac{3}{2}} + \frac{1}{2} [\frac{2 x + 1}{4} \sqrt{{(x + \frac{1}{2})}^{2} + \frac{7}{4}} + \frac{7}{8}] + c$

$\frac{1}{3} {(x^{2} + x + 2)}^{\frac{3}{2}} + \frac{1}{2} \sin^{- 1} (\frac{x + \frac{1}{2}}{\frac{\sqrt{7}}{2}}) + c$

$\frac{1}{3} {(x^{2} + x + 2)}^{\frac{3}{2}} + \frac{1}{2} [\frac{2 x + 1}{4} \sqrt{{(x + \frac{1}{2})}^{2} + \frac{7}{4}} + \frac{7}{8} \sin^{- 1} (\frac{2 x + 1}{\sqrt{7}})] + c$

answer is D.

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Detailed Solution

$\begin{array}{l} Let x + 1 = A \frac{d}{d x} (x^{2} + x + 2) + B \\ x + 1 = A (2 x + 1) + B \\ x + 1 = 2 A x + A + B \\ On comparing \\ 2 A = 1 A + B = 1 \\ A = \frac{1}{2} B = \frac{1}{2} \\ ∴ \int (x + 1) \sqrt{x^{2} + x + 2} d x \end{array}$

$\begin{array}{l} = \int (\frac{1}{2} (x + 1) + \frac{1}{2}) \sqrt{x^{2} + x + 2} d x \\ = \frac{1}{2} \int (2 x + 1) \sqrt{x^{2} + x + 2} d x + \frac{1}{2} \sqrt{x^{2} + x + 2} d x \\ = (∴ \int f^{'} (x) \sqrt{f (x)} d x = \frac{2}{3} f^{\frac{3}{2}} (x) + c) \end{array}$

$\begin{array}{l} = \frac{1}{2} \cdot \frac{2}{3} {(x^{2} + x + 2)}^{\frac{3}{2}} + \frac{1}{2} \int \sqrt{(x^{2} + 2 x \cdot \frac{1}{2} + \frac{1}{4} - \frac{1}{4} + 2)} d x \\ = \frac{1}{3} {(x^{2} + x + 2)}^{\frac{3}{2}} + \frac{1}{2} \int \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{7}}{2})}^{2}} d x \\ = \frac{1}{3} {(x^{2} + x + 2)}^{\frac{3}{2}} + \frac{1}{2} [\frac{(x + \frac{1}{2})}{2} \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{7}}{2})}^{2}} + \frac{{(\frac{\sqrt{7}}{2})}^{2}}{2} \sin x^{- 1} (\frac{x + \frac{1}{2}}{\frac{\sqrt{7}}{2}})] + c \\ = \frac{1}{3} {(x^{2} + x + 2)}^{\frac{3}{2}} + \frac{1}{2} [\frac{(2 x + 1)}{4} \sqrt{{(x + \frac{1}{2})}^{2} + \frac{7}{4}} + \frac{7}{8} \sin x^{- 1} (\frac{2 x + 1}{\sqrt{7}})] + c \end{array}$