int frac{d x}{x ln (x) ln ^2(x) ln ^3(x) ldot | vedclass

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(D) Given the integral: $I = \int \frac{d x}{x \ln (x) \ln ^2(x) \ldots \ln ^m(x)}$Let $u = \ln(x)$, then $du = \frac{1}{x} dx$.The integral becomes:

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$(m+2)(1-m)$

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(D) Given the integral: $I = \int \frac{d x}{x \ln (x) \ln ^2(x) \ldots \ln ^m(x)}$Let $u = \ln(x)$, then $du = \frac{1}{x} dx$.The integral becomes: $I = \int \frac{du}{u^1 \cdot u^2 \cdot u^3 \cdot \ldots \cdot u^m} = \int \frac{du}{u^{1+2+3+\ldots+m}}$.Using the sum formula $\sum_{i=1}^{m} i = \frac{m(m+1)}{2}$, we have:$I = \int u^{-\frac{m(m+1)}{2}} du$.Applying the power rule $\int u^n du = \frac{u^{n+1}}{n+1} + C$:$I = \frac{u^{-\frac{m(m+1)}{2} + 1}}{-\frac{m(m+1)}{2} + 1} + C = \frac{u^{\frac{2 - m^2 - m}{2}}}{\frac{2 - m^2 - m}{2}} + C$.Factorizing the exponent: $2 - m^2 - m = -(m^2 + m - 2) = -(m+2)(m-1) = (m+2)(1-m)$.So, $I = \frac{u^{\frac{(m+2)(1-m)}{2}}}{\frac{(m+2)(1-m)}{2}} + C$.Comparing this with the given form $\frac{(\ln x)^K}{K} + C$, we identify $K = \frac{(m+2)(1-m)}{2}$.Therefore, $2K = (m+2)(1-m)$.

$\int \frac{d x}{x \ln (x) \ln ^2(x) \ln ^3(x) \ldots \ln ^m(x)}=\frac{(\ln (x))^K}{K}+C$
$\Rightarrow 2 K=$

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$\int \frac{d x}{x \ln (x) \ln ^2(x) \ln ^3(x) \ldots \ln ^m(x)}=\frac{(\ln (x))^K}{K}+C$$\Rightarrow 2 K=$