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 -

Options

A. `I_(1000) + 999 I_(998)`
B. `I_(890) + 890 I_(889)`
C. `I_(100) + 100 I_(99)`
D. `I_(53) + 54 I_(52)`

Answer

C

Explanation

`I_(n)=overset(e)underset(1)intunderset(II)(1).underset(I)((logx)^(n))dx`
` I_(n) (logx)^(n)x|_(1)^(e)-overset(e)underset(1)int(n(logx)^(n-1))/(x).xdx`
`I_(n) = e - 0 - nI_(n-1)`
`I_(n) + nI_(n-1) = e`
`I_(2001+2011)I_(2010) = e`
`I_(100) + 100I_(99) = e`