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

Integrals

Solution:

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