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

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Question

(C) Let $I = \int \frac{(x + 1)(x + \log x)^2}{x} \, dx$. We can rewrite the integrand as $\int (x + \log x)^2 \left( \frac{x + 1}{x} \right) \, dx =

Vedclass · Accepted Answer

$\frac{1}{3}(x + \log x)^3 + c$

Vedclass · Answer

(C) Let $I = \int \frac{(x + 1)(x + \log x)^2}{x} \, dx$. We can rewrite the integrand as $\int (x + \log x)^2 \left( \frac{x + 1}{x} \right) \, dx = \int (x + \log x)^2 \left( 1 + \frac{1}{x} \right) \, dx$. Let $t = x + \log x$. Then,differentiating with respect to $x$,we get $dt = (1 + \frac{1}{x}) \, dx$. Substituting these into the integral,we get $I = \int t^2 \, dt$. Integrating $t^2$ with respect to $t$,we get $I = \frac{t^3}{3} + c$. Substituting back $t = x + \log x$,we get $I = \frac{1}{3}(x + \log x)^3 + c$.

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

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