Solve I_n + n I_{n-1},if I_n = int (ln x)^n d | vedclass

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(B) Given $I_n = \int (\ln x)^n dx$. Using integration by parts,let $u = (\ln x)^n$ and $dv = dx$. Then $du = n(\ln x)^{n-1} \cdot \frac{1}{x} dx$ and

Vedclass · Accepted Answer

$x(\ln x)^n + k$

Vedclass · Answer

(B) Given $I_n = \int (\ln x)^n dx$. Using integration by parts,let $u = (\ln x)^n$ and $dv = dx$. Then $du = n(\ln x)^{n-1} \cdot \frac{1}{x} dx$ and $v = x$. $I_n = x(\ln x)^n - \int x \cdot \frac{n(\ln x)^{n-1}}{x} dx + k$. $I_n = x(\ln x)^n - n \int (\ln x)^{n-1} dx + k$. Since $I_{n-1} = \int (\ln x)^{n-1} dx$,we have: $I_n = x(\ln x)^n - n I_{n-1} + k$. Rearranging the terms,we get: $I_n + n I_{n-1} = x(\ln x)^n + k$.

Solve $I_n + n I_{n-1}$,if $I_n = \int (\ln x)^n dx$.

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