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Q.

∫131(tan−1x)31+x2 dx=

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a

14(π12)6

b

14(π12)4

c

14((π4)4−(π6)4)

d

(π8)4

answer is C.

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Detailed Solution

∫π/6π/4t3.dt                           Tan−1x=t =(t44)π/6π/4 =14(π444−π464)
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