∫1(2x+1)x2−x−2dx=

− 1 5 tan − 1 ⁡ 4 x + 7 6 x + 3 + c

− 1 5 tan − 1 ⁡ 4 x + 7 6 x + 3 + c

− 1 5 cos − 1 ⁡ 4 x + 7 6 x + 3 + c

− 1 5 sin − 1 ⁡ 4 x − 7 6 x − 3 + c

− 1 5 sin − 1 ⁡ 4 x + 7 6 x + 3 + c

− 1 5 tan − 1 ⁡ 4 x + 7 6 x + 3 + c

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

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$- \frac{1}{\sqrt{5}} \cos^{- 1} (\frac{4 x + 7}{6 x + 3}) + c$

$- \frac{1}{\sqrt{5}} \sin^{- 1} (\frac{4 x - 7}{6 x - 3}) + c$

$- \frac{1}{\sqrt{5}} \sin^{- 1} (\frac{4 x + 7}{6 x + 3}) + c$

$- \frac{1}{\sqrt{5}} \tan^{- 1} (\frac{4 x + 7}{6 x + 3}) + c$

answer is D.

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$\int \frac{1}{(2 x + 1) \sqrt{x^{2} - x - 2}} dx p u t 2 x + 1 = \frac{1}{t} \Rightarrow 2 d x = \frac{- 1}{t^{2}} d t$

$\int \frac{1}{\frac{1}{t} \sqrt{{(\frac{1 - t}{2 t})}^{2} - (\frac{1 - t}{2 t}) - 2}} (\frac{- 1}{2 t^{2}} dt)$

$\int \frac{- dt}{\sqrt{t^{2} - 2 t + 1 - 2 t + 2 t^{2} - 8 t^{2}}}$

$= \int \frac{- dt}{\sqrt{- 5 t^{2} - 4 t + 1}}$

$= \frac{- dt}{\sqrt{5 [{(\frac{3}{5})}^{2} - {(t + \frac{2}{5})}^{2}}}$

$= \frac{- 1}{\sqrt{5}} [\sin^{1} (\frac{5 t + 2}{3})] + c$

$= - \frac{1}{\sqrt{5}} \sin^{- 1} (\frac{\frac{5}{2 x + 1}}{3} + 2)$

$= - \frac{1}{\sqrt{5}} \sin^{- 1} (\frac{4 x + 7}{6 x + 3}) + c$

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