int frac{1}{x(log x)^2} , dx = | vedclass

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Question

(B) To solve the integral $\int \frac{1}{x(\log x)^2} \, dx$,we use the method of substitution.Let $t = \log x$.Then,differentiating both sides with r

Vedclass · Accepted Answer

$-\frac{1}{\log x} + c$

Vedclass · Answer

(B) To solve the integral $\int \frac{1}{x(\log x)^2} \, dx$,we use the method of substitution.Let $t = \log x$.Then,differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{x}$,which implies $dt = \frac{1}{x} \, dx$.Substituting these into the integral,we get:$\int \frac{1}{t^2} \, dt = \int t^{-2} \, dt$.Using the power rule for integration $\int x^n \, dx = \frac{x^{n+1}}{n+1} + c$,we have:$\int t^{-2} \, dt = \frac{t^{-2+1}}{-2+1} + c = \frac{t^{-1}}{-1} + c = -\frac{1}{t} + c$.Finally,substituting $t = \log x$ back into the expression,we get:$-\frac{1}{\log x} + c$.

$\int \frac{1}{x(\log x)^2} \, dx = $

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