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

∫1x(log⁡x)2−2log⁡x+102dxlog⁡x=t⇒1xdx=dtI=∫1(t−1)2+92dt Put t−1=3tan⁡θdt=3sec2⁡θdθI=∫3sec2⁡θ9sec2⁡θ2dθ=127∫cos2⁡θdθ=127∫12+12cos⁡2θdθ=154θ+sin⁡2θ2+C=154tan−1⁡t−13+122tan⁡θ1+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