x-1=t2⇒x=t2+12I=∫03tan-1tdt2+12=tan-1t·t2+1203-∫03t2+1dt=16π3-t33+t03=16π3-23.

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$\int_{1}^{16} \tan^{- 1} \sqrt{\sqrt{x} - 1} d x =$

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$\frac{8 π}{3} + \sqrt{3}$

$\frac{8 π}{3} - \sqrt{3}$

$\frac{16 π}{3} - 2 \sqrt{3}$

$\frac{16 π}{3} + 2 \sqrt{3}$

answer is C.

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Detailed Solution

$\sqrt{x} - 1 = t^{2} \Rightarrow x = {(t^{2} + 1)}^{2}$

$I = \int_{0}^{\sqrt{3}} \tan^{- 1} t d {(t^{2} + 1)}^{2} = {\tan^{- 1} t \cdot {(t^{2} + 1)}^{2}|}_{0}^{\sqrt{3}} - \int_{0}^{\sqrt{3}} (t^{2} + 1) d t$

$= \frac{16 π}{3} - {(\frac{t^{3}}{3} + t)}_{0}^{\sqrt{3}} = \frac{16 π}{3} - 2 \sqrt{3}$ .

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