∫2x+tan−1x1+x2dx=

I=∫2x+tan−1⁡x1+x2dxI=∫2x1+x2dx+∫tan−1⁡x1+x2dx=∫2x1+x2dx+∫tdt ∵tan−1⁡x=t,11+x2dx=dt=log⁡1+x2+23t32+C=log⁡1+x2+23tan−1⁡x32+C

∫2x+tan−1x1+x2dx=

I=∫2x+tan−1⁡x1+x2dxI=∫2x1+x2dx+∫tan−1⁡x1+x2dx=∫2x1+x2dx+∫tdt ∵tan−1⁡x=t,11+x2dx=dt=log⁡1+x2+23t32+C=log⁡1+x2+23tan−1⁡x32+C

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

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$\log (1 + x^{2})$

$\log (1 + x^{2}) + {(\tan^{- 1} x)}^{\frac{3}{2}} + C$

$\log (1 + x^{2}) + {(\cot^{- 1} x)}^{\frac{3}{2}} + C$

$\log (1 + x^{2}) + \frac{2}{3} {(t a n^{- 1} x)}^{\frac{3}{2}} + C$

answer is D.

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Detailed Solution

$\begin{array}{l} I = \int \frac{2 x + \sqrt{\tan^{- 1} x}}{1 + x^{2}} d x \\ I = \int \frac{2 x}{1 + x^{2}} d x + \int \frac{\sqrt{\tan^{- 1} x}}{1 + x^{2}} d x \\ = \int \frac{2 x}{1 + x^{2}} d x + \int \sqrt{t} d t ∵ \tan^{- 1} x = t, \frac{1}{1 + x^{2}} d x = d t \\ = \log (1 + x^{2}) + \frac{2}{3} t^{\frac{3}{2}} + C \\ = \log (1 + x^{2}) + \frac{2}{3} {(\tan^{- 1} x)}^{\frac{3}{2}} + C \end{array}$