Question

$\int \frac{1}{x}\left\{{\log }_{ex}e\cdot {\log }_{e^{2}x}e\cdot {\log }_{e^{3}x}e\right\}dx$ is equal to

• A. $\frac{1}{2}\log \left\{{\log }_e{e}^{2}x\right\}-\log \left\{{\log }_e{e}^{3}x\right\}+\log \left\{{\log }_e{e}^{4}x\right\}+C$
• B. $\frac{1}{2}\log \left\{{\log }_{e}x\right\}-\log \left\{{\log }_{e}x\right\}+\log \left\{{\log }_{e}x\right\}+C$
• C. $\frac{1}{2}\log \left\{{\log }_{e}ex\right\}-\log \left\{{\log }_e{e}^2{x}^2\right\}+\log \left\{{\log }_e{e}^3{x}^3\right\}+C$
• D. $\frac{1}{2}\log \left\{{\log }_{e}ex\right\}-\log \left\{{\log }_e{e}^{2}x\right\}+\frac{1}{2}\log \left\{{\log }_e{e}^{3}x\right\}+C$

Answer 1

Answer: (D)

We have,
$\int \frac{1}{x}\left[{\log }_{ex}e\cdot {\log }_{e^{2}x}\cdot {\log }_{e^{3}x}e\right]dx$
$=\int \frac{1}{x{\log }_{e}ex{\log }_e{e}^{2}x\cdot {\log }_e{e}^{3}x}dx$
$=\int \frac{1}{x\left({\log }_{e}e+{\log }_{e}x\right)\left({\log }_e{e}^2+{\log }_{e}x\right)}\left({\log }_e{e}^3+{\log }_{e}x\right)dx$
$=\int \frac{d\left({\log }_{e}x\right)}{\left(1+{\log }_{e}x\right)\left(2+{\log }_{e}x\right)\left(3+{\log }_{e}x\right)}$
$=\int \frac{1}{(1+t)(2+t)(3+t)}dt$, where $t={\log }_{e}x$
$=\int \left(\frac{1}{2}\cdot \frac{1}{1+t}-\frac{1}{2+t}+\frac{1}{2}\cdot \frac{1}{3+t}\right)dt$
$=\frac{1}{2}\log \left|1+{\log }_{e}x\right|-\log \left|2+{\log }_{e}x\right|$
$+\frac{1}{2}\log \left|3+{\log }_{e}x\right|+C$
$=\frac{1}{2}\log \left\{{\log }_{e}ex\right\}-\log \left\{{\log }_e{e}^{2}x\right\}+\frac{1}{2}\log \left\{{\log }_e{e}^{3}x\right\}+C$