The number of solutions of the equation 6 int | vedclass

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(B) Let $u = |x|$. Since $x \in R \setminus \{0\}$,we have $u > 0$. The equation is $6 \int_{0}^{u} (t^2-1) \ln t \ dt = 5u$.Using integration by part

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$4$

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(B) Let $u = |x|$. Since $x \in R \setminus \{0\}$,we have $u > 0$. The equation is $6 \int_{0}^{u} (t^2-1) \ln t \ dt = 5u$.Using integration by parts,$\int (t^2-1) \ln t \ dt = \ln t (\frac{t^3}{3} - t) - \int \frac{1}{t} (\frac{t^3}{3} - t) dt = \ln t (\frac{t^3}{3} - t) - (\frac{t^3}{9} - t) + C$.Evaluating the definite integral from $0$ to $u$ (taking the limit as $t 	o 0^+$): $\lim_{t 	o 0^+} [\ln t (\frac{t^3}{3} - t) - \frac{t^3}{9} + t] = 0$.So,$6 [\ln u (\frac{u^3}{3} - u) - \frac{u^3}{9} + u] = 5u$.Dividing by $u$ (since $u > 0$): $6 [\ln u (\frac{u^2}{3} - 1) - \frac{u^2}{9} + 1] = 5$.$2 \ln u (u^2 - 3) - \frac{2u^2}{3} + 6 = 5 \implies 2 \ln u (u^2 - 3) = \frac{2u^2}{3} - 1$.$\ln u = \frac{2u^2 - 3}{6(u^2 - 3)}$.Let $f(u) = \ln u$ and $g(u) = \frac{2u^2 - 3}{6(u^2 - 3)}$. By analyzing the graphs of $f(u)$ and $g(u)$ for $u > 0$,we find two intersection points for $u$. Since $u = |x|$,each positive value of $u$ gives two values of $x$ $(x = \pm u)$. Thus,there are $2 	imes 2 = 4$ solutions.

The number of solutions of the equation $6 \int_{0}^{|x|} ((t^2-1) \ln t) dt = 5|x|$ for $x \in R \setminus \{0\}$ is

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