If In - - ln xn dx then In - nIn-1

I_n=x(ln x)^n ∫x(n)(ln x)^n 1/xdx =x(ln x)^n nI_(n 1) ⇒ I_n+nI_n 1=x(ln x)^n ...

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Integration by Parts

If In = ∫ ln ...

Question

If $I_{n} = \int (l n x)^{n} d x t h e n I_{n} + n I_{n - 1}$

$\frac{(l n x)^{n}}{x}$

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$x (l n x)^{n - 1}$

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$x (l n x)^{n}$

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None of these

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Solution

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Similar questions

If $I_{n} = \int (l n x)^{n} d x t h e n I_{n} + n I_{n - 1}$

I_{n} = \int_{0}^{\infty} x^{n} e^{- x} d x (n

) = n!

I_{n} = n I_{n - 1}

I_{n} = \int (ln x)^{n} d x,

I_{n} + n I_{n - 1} =

I_{n} = \frac{d^{n}}{d x^{n}} (x^{n} {log}_{e} x)

I_{n} - n I_{n - 1} =

n ϵ N

I_{n} = \int (log x)^{n} d x

I_{n} + n I_{n + 1} =

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