Question

If $\int [\log \left(\log x\right)+\frac{1}{{\left(\log x\right)}^2}]dx=x\left[f\right(x)-g(x\left)\right]+c$ then

• A. $f\left(x\right)=\log \left(\log x\right);g\left(x\right)=\frac{1}{\log x}$
• B. $f\left(x\right)=\log x;g\left(x\right)=\frac{1}{\log x}$
• C. $f\left(x\right)=;g\left(x\right)=\frac{1}{\log x};g\left(x\right)=\log \left(\log x\right)$
• D. $f\left(x\right)=\frac{1}{x\log x};g\left(x\right)=\frac{1}{\log x}$

Answer 1

The correct option is A $f\left(x\right)=\log \left(\log x\right);g\left(x\right)=\frac{1}{\log x}$

Let $I=\int \{\log \left(\log x\right)+\frac{1}{{\left(\log x\right)}^2}\}dx$
Put $\log x=t$ or $x=e^t$ $dx=e^{t}dt$
$\therefore I=\int (\log t+\frac{1}{t^2})e^{t}dt=\int (\log t+\frac{1}{t}-\frac{1}{t}+\frac{1}{t^2})dt$
$=\int (\log t+\frac{1}{t})e^{t}dt+\int (-\frac{1}{t}+\frac{1}{t^2})e^{t}dt$
$=\int \log t{e}^{t}dt+\int \frac{e^t}{t}dt+\int (-\frac{1}{t})e^{t}dt+\int \left(e^t\frac{1}{t^2}\right)dt$
Integrating by parts we get
$=\left(\log t\right)e^t-\int \frac{1}{t}{e}^{t}dt+\int e^t.\frac{1}{t}dt+\int (-\frac{1}{t})e^t-\int \frac{1}{t^2}{e}^{t}dt+\int e^t\frac{1}{t^2}dt+c=x(\log \log x-\frac{1}{\log x})+c=x\left(f\right(x)-g(x\left)\right)+c$
$\therefore f\left(x\right)\log \log x;g\left(x\right)=\frac{1}{\log x}$