If In=int( lnx)^n dx then In+nI(n-1)...

If `I_n=int( lnx)^n dx` then `I_n+nI_(n-1)`

`(("In "x)^(n))/(x)+C`

`x("In "x)^(n-1)+C`

`x("In "x)^(n)+C`

none of these

Text Solution

Verified by Experts

The correct Answer is:

`I_(n)=x("In "x)^(n)-int (x(n)("In "x)^(n-1))/(x)dx+C=x ("In "x)^(n)-n I_((n-1))+C`
or `I_(n)+nI_(n-1)=x("In "x)^(n)+C`

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