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Question is attached in image ….jitendra Sir explain this solution. I will follow your method but I want this too In = dn-1/dxn-1 [xn-1 + nxn-1 log x ] In = (n-1) dn-2 / dx n-2 xn-2 + n dn-1 / dxn-1 (xn-1 logx) In =(n-1 ) ! + nIn-1 In – nIn-1 =(n-1)! Proved

Question is attached in image ….jitendra Sir explain this solution.  I will follow your method but I want this too                           In = dn-1/dxn-1 [xn-1  + nxn-1 log x ]                In = (n-1) dn-2 / dx n-2 xn-2 + n dn-1 / dxn-1 (xn-1 logx) 
In =(n-1 ) ! + nIn-1 
In – nIn-1 =(n-1)! Proved  

Grade:12

4 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Hello student,
Please find answer to your question
I_{n} = \frac{d^{n}}{dx^{n}}(x^{n}log(x))
I_{n} = \frac{d^{n-1}}{dx^{n-1}}(x^{n}.\frac{1}{x}+log(x).nx^{n-1})
I_{n} = \frac{d^{n-1}}{dx^{n-1}}(x^{n-1}+nx^{n-1}log(x))
I_{n} = (n-1)\frac{d^{n-1}}{dx^{n-1}}(x^{n-2})+n\frac{d^{n-1}}{dx^{n-1}}(x^{n-1}log(x))
I_{n} = (n-1)!+n\frac{d^{n-1}}{dx^{n-1}}(x^{n-1}log(x))…........(1)
I_{n-1} = \frac{d^{n-1}}{dx^{n-1}}(x^{n-1}log(x))
Put this in equation (1), we have
I_{n} = (n-1)! + nI_{n-1}
I_{n} - nI_{n-1} = (n-1)!
milind
23 Points
9 years ago
Jitendra Sir Can u explain me second step ....
milind
23 Points
9 years ago
Second of answer ….please explain …..please 
milind
23 Points
9 years ago
Second step of answer please explain 

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