Question

Find the antiderivative of $f\left(x\right)=ln\left(lnx\right)+{\left(lnx\right)}^{-2}$ whose graph passes through $(e,e)$.

Answer 1

$\begin{array}{c}Weget\\ f\left(x\right)=\ln \left(\ln x\right)+\frac{1}{{\left(\ln x\right)}^2}\\ \int f\left(x\right)dx=\int \left(\ln \left(\ln x\right)+\frac{1}{{\left(\ln x\right)}^2}\right)dx\\ x=e^t,dx=e^{t}dt\\ =\int \left(\ln t+\frac{1}{t^2}\right)e^{t}dt\\ =\int e^t\left(lnt+\frac{1}{t}\right)dt+\int e^t\left(\frac{1}{t^2}-\frac{1}{t}\right)dt\\ =\int e^t\left(\ln t+\frac{d\left(\ln t\right)}{dt}\right)dt+\int e^t\left(-\frac{1}{t}+\frac{d}{dt}\left(\frac{-1}{t}\right)dt\right)\\ =e^t\ln t-\frac{e^t}{t}+C\\ \int fdx=x\ln \left(\ln x\right)-\frac{x}{\ln x}+C\\ y=x\ln \left(\ln x\right)\frac{-e}{1}+cput\left(e,e\right)\\ e=e\ln \left(1\right)-\frac{e}{1}+C\\ c=2e\\ y=x\ln \left(lnx\right)\frac{-x}{\ln x}+2e\end{array}$