int frac{1}{xleft[6(log x)^2+7 log x+2right]} | vedclass

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(B) $I = \int \frac{1}{x[6(\log x)^2 + 7 \log x + 2]} dx$Let $t = \log x$,then $dt = \frac{1}{x} dx$.Substituting these into the integral,we get:$I =

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$\log \left|\frac{2 \log x+1}{3 \log x+2}\right|+C$

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(B) $I = \int \frac{1}{x[6(\log x)^2 + 7 \log x + 2]} dx$Let $t = \log x$,then $dt = \frac{1}{x} dx$.Substituting these into the integral,we get:$I = \int \frac{dt}{6t^2 + 7t + 2}$Factorizing the denominator: $6t^2 + 7t + 2 = 6t^2 + 4t + 3t + 2 = 2t(3t + 2) + 1(3t + 2) = (2t + 1)(3t + 2)$.Using partial fractions: $\frac{1}{(3t + 2)(2t + 1)} = \frac{A}{3t + 2} + \frac{B}{2t + 1}$$1 = A(2t + 1) + B(3t + 2)$For $t = -\frac{1}{2}$,$1 = B(3(-\frac{1}{2}) + 2) = B(-\frac{3}{2} + 2) = B(\frac{1}{2}) \implies B = 2$.For $t = -\frac{2}{3}$,$1 = A(2(-\frac{2}{3}) + 1) = A(-\frac{4}{3} + 1) = A(-\frac{1}{3}) \implies A = -3$.Thus,$I = \int (\frac{-3}{3t + 2} + \frac{2}{2t + 1}) dt$$I = -\frac{3 \log |3t + 2|}{3} + \frac{2 \log |2t + 1|}{2} + C$$I = -\log |3t + 2| + \log |2t + 1| + C$$I = \log |\frac{2t + 1}{3t + 2}| + C$Substituting $t = \log x$ back,we get:$I = \log |\frac{2 \log x + 1}{3 \log x + 2}| + C$

$\int \frac{1}{x\left[6(\log x)^2+7 \log x+2\right]} d x$ is

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