Q115 / 160 · TS EAMCET 2013 · Mathematics
If $\int \frac{d x}{x(\log x-2)(\log x-3)}=I+C$, then $I$ is equal to
- A$\frac{1}{x} \log \left|\frac{\log x-3}{\log x-2}\right|$
- Correct answerB$\log \left|\frac{\log x-3}{\log x-2}\right|$
- C$\log \left|\frac{\log x-2}{\log x-3}\right|$
- D$\log |(\log x-3)(\log x-2)|$
Solution
Let $I_1=\int \frac{d x}{x(\log x-2)(\log x-3)}$
Let $t=\log x \Rightarrow d t=\frac{d x}{x}$
$$
\begin{aligned}
\therefore \quad I_1 & =\int \frac{d t}{(t-2)(t-3)} \\
& =\int\left[\frac{1}{(t-3)}-\frac{1}{(t-2)}\right] d t \\
& =\log |(t-3)|-\log |(t-2)|+C \\
& =\log \left|\frac{\log x-3}{\log x-2}\right|+C \\
\Rightarrow \quad I_1 & =\int \frac{d x}{x(\log x-2)(\log x-3)} \\
& =\log \left|\frac{\log x-3}{\log x-2}\right|+C
\end{aligned}
$$
But given, $\int \frac{d x}{x(\log x-2)(\log x-3)}=I+C$
On comparing, we get
$$
I=\log \left|\frac{\log x-3}{\log x-1}\right|
$$
Hence, the correct answer is option: 'B'.