Question

The anti-derivative of $f\left(x\right)=\log \left(\log x\right)+(\log x{)}^{-2}$ whose graph passes through $(e,e)$ is

• A. $x\left[\log \right(\log x)+(\log x{)}^{-1}]$
• B. $x[-\log (\log x)+(\log x{)}^{-1}]+e$
• C. $x\left[\log \right(\log x)-(\log x{)}^{-1}]+2e$
• D. $x\left[\log \right(\log x)-(\log x{)}^{-1}]+3e$

Answer 1

Answer: (C)

$x\left[\log \right(\log x)-(\log x{)}^{-1}]+2e$

We need to find;

$I=\int f\left(x\right)dxI=\int \log \left(\log x\right)+\frac{1}{{\left(\log x\right)}^2}dxletx=e^t\therefore dx=e^{t}dt\&\log x=tI=\int (\log t+\frac{1}{t^2})e^{t}dtI=\int (\log t+\frac{1}{t}-\frac{1}{t}+\frac{1}{t^2})e^{t}dtI=\int (\log t+\frac{1}{t})e^{t}dt-\int (\frac{1}{t}-\frac{1}{t^2})e^{t}dt$

If we look carefully, the two integrals are in the form of;

$\int (f\left(x\right)+f^{{}^{\prime }}\left(x\right))e^{x}dx$

And this above integral as we know equates to;

$\int (f\left(x\right)+f^{{}^{\prime }}\left(x\right))e^{x}dx=e^{x}f\left(x\right)+C$

hence;

$I=\int (\log t+\frac{1}{t})e^{t}dt-\int (\frac{1}{t}-\frac{1}{t^2})e^{t}dt\log t=f\left(t\right)\&f^{{}^{\prime }}\left(t\right)=\frac{1}{t}\frac{1}{t}=g\left(t\right)\&g^{{}^{\prime }}\left(t\right)=-\frac{1}{t^2}\therefore I=\int (f\left(t\right)+f^{{}^{\prime }}\left(t\right))e^{t}dt-\int (g\left(t\right)+g^{{}^{\prime }}\left(t\right))e^{t}dtI=f\left(t\right)e^t-g\left(t\right)e^t+CI=(\log t-\frac{1}{t})e^t+CI\left(x\right)=(\log \left(\log x\right)-\frac{1}{\log x})x+C$

Now we know that the graph passes thru $(e,e)$, hence;

$I\left(e\right)=e$

$I\left(e\right)=(\log \left(\log e\right)-\frac{1}{\log e})e+C$

$I\left(e\right)=(\log \left(1\right)-\frac{1}{1})e+C=e$

$-e+C=e$

$\therefore C=2e$

Hence the antiderivate of $f\left(x\right)$ is;

$\int f\left(x\right)dx=x(\log \left(\log x\right)-\frac{1}{\log x})+2e$