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Q.

∫2x+tan−1x1+x2dx=

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a

log(1+x2)

b

log(1+x2)+tan−1x32+C

c

log(1+x2)+cot−1x32+C

d

log(1+x2)+23tan−1x32+C

answer is D.

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Detailed Solution

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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