If ∫$$1x(log$$⁡$$x)2$$−$$2log$$⁡$$x+102dx=154tan$$−$$1$$⁡$$(f(x))+3(log$$⁡x−$$1)g(x)+C$$ then $$f(e)+g(e)=$$−−−

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If ∫$$1x(log$$⁡$$x)2$$−$$2log$$⁡$$x+102dx=154tan$$−$$1$$⁡$$(f(x))+3(log$$⁡x−$$1)g(x)+C$$ then $$f(e)+g(e)=$$−−−

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∫$$1x(log$$⁡$$x)2$$−$$2log$$⁡$$x+102dxlog$$⁡$$x=t$$⇒$$1xdx=dtI=$$∫$$1(t$$−$$1)2+92dt$$ Put t−$$1=3tan$$⁡θ$$dt=3sec2$$⁡θdθ$$I=$$∫$$3sec2$$⁡θ$$9sec2$$⁡θ$$2d$$θ$$=127$$∫$$cos2$$⁡θdθ$$=127$$∫$$12+12cos$$⁡$$2$$θdθ$$=154$$θ$$+$$$$sin$$⁡$$2$$θ$$2+C=154tan$$−$$1$$⁡t−$$13+122tan$$⁡θ$$1+tan2$$⁡θ$$+C=154tan$$−$$1$$⁡t−$$13+3(t$$−$$1)t2$$−$$2t+10+Cf(x)=log$$⁡x−$$13$$,$$g(x)=(log$$⁡$$x)2$$−$$2log$$⁡$$x+10$$

$If \int \frac{1}{x {((\log x)^{2} - 2 \log x + 10)}^{2}} d x = \frac{1}{54} [\tan^{- 1} (f (x)) + \frac{3 (\log x - 1)}{g (x)}] + C then f (e) + g (e) = - - -$

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If ∫1x(log⁡x)2−2log⁡x+102dx=154tan−1⁡(f(x))+3(log⁡x−1)g(x)+C then f(e)+g(e)=−−−

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$If \int \frac{1}{x {((\log x)^{2} - 2 \log x + 10)}^{2}} d x = \frac{1}{54} [\tan^{- 1} (f (x)) + \frac{3 (\log x - 1)}{g (x)}] + C then f (e) + g (e) = - - -$