Let I(n) = int(0)^(1) (logx)^(n)dx, wher...

Let `I_(n) = int_(0)^(1) (logx)^(n)dx`, where n is a non-negative integer. Then `I_(2001) + 2011 I_(2010)` is equal to -

A

`I_1000 + 999 I_998`

B

`I_90 + 890 I_889`

C

`I_100 + 100 I_99`

D

`I_13 + 54 I_52`

Verified by Experts

The correct Answer is:

C

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