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milind , 11 Years ago

Grade 12

2 Answers

Jitender Singh

Hello student,
Please find answer to your question
$I_{n} = \frac{d^{n}}{dx^{n}}(x^{n}log(x))$
$I_{1} = \frac{d}{dx}(xlog(x))$
$I_{1} = 1 + log(x)$
$I_{2} = \frac{d^{2}}{dx^{2}}(x^{2}log(x))$
$I_{2} = 3 + 2log(x)$
$I_{3} = \frac{d^{3}}{dx^{3}}(x^{3}log(x))$
$I_{3} = 11 + 6log(x)$
$I_{4} = \frac{d^{4}}{dx^{4}}(x^{4}log(x))$
$I_{4} = 50 + 24log(x)$
$I_{5} = \frac{d^{5}}{dx^{5}}(x^{5}log(x))$
$I_{5} = 274 + 120log(x)$
$I_{5}-5I_{4} = 24 = 4! = (5-1)!$
$I_{4}-4I_{3} = 6 = 3! = (4-1)!$
$I_{3}-3I_{2} = 2 = 2! = (3-1)!$
Similarly, you can prove for n,
$I_{n}-nI_{n-1} = (n-1)!$

Approved

Last Activity: 11 Years ago

milind

Sir explain this solution. I_{n =}d^n-1/dx^n-1[x^n-1 + nx^n-1 log x ] In = (n-1) d^n-2/ dx ^n-2x^n-2+ n d^n-1/ dx^n-1(x^n-1logx)

I_{n =}(n-1 ) ! + nI_n-1

In – nI_n-1=(n-1)! Proved

Last Activity: 11 Years ago

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