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Q.

∫1x(1+logx)3 dx=

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a

−  12(1+logx)2+c

b

−12(1+logx)3+c

c

11+logx+c

d

−  13(1+logx)3+c

answer is A.

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Detailed Solution

1+logx=t                  1x dx=dt       ∫1x(1+logx)3 dx=       ∫1t3 dt=−12t2+c                           =−121(1+logx)2+c
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