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ADNAN MUHAMMED

Grade 12,

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Solve this question sir please

Solve this question sir please 

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Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
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}) + 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 (1), we have
I_{n} =(n-1)! + nI_{n-1}
I_{n} - nI_{n-1} = (n-1)!

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