Q100 / 180 · KCET 2024 · Mathematics
$\int \frac{1}{x\left[6(\log x)^2+7 \log x+2\right]} d x$ is
- A$\frac{1}{2} \log \left|\frac{2 \log x+1}{3 \log x+2}\right|+C$
- Correct answerB$\log \left|\frac{2 \log x+1}{3 \log x+2}\right|+C$
- C$\log \left|\frac{3 \log x+2}{2 \log x+1}\right|+C$
- D$\frac{1}{2} \log \left|\frac{3 \log x+2}{2 \log x+1}\right|+C$
Solution
$I=\int \frac{1}{x\left[6(\log x)^2+7 \log x+2\right]} d x$
$=\int \frac{d t}{6 t^2+7 t+2} \quad\left[\begin{array}{l}\because \text { put } t=\log x \\ d t=\frac{1}{x} d x\end{array}\right]$
$=\int \frac{d t}{(3 t+2)(2 t+1)}$
Now, $\frac{1}{(3 t+2)(2 t+1)}=\frac{A}{(3 t+2)}+\frac{B}{(2 t+1)}$
$=\frac{A(2 t+1)+B(3 t+2)}{(3 t+2)(2 t+1)}$
$\Rightarrow \quad 1=A(2 t+1)+B(3 t+2)$
Now, on putting $t=\frac{-1}{2}$
$\Rightarrow \quad 1=0+B\left[3\left(-\frac{1}{2}\right)+2\right]$
$\Rightarrow \quad B=2$
and put $t=\frac{-2}{3}$
$\Rightarrow \quad 1=A\left(2 \times\left(-\frac{2}{3}\right)+1\right)+0 \Rightarrow 1=A\left(-\frac{4}{3}+1\right)$
$\Rightarrow \quad A=-3$
$\therefore \quad I=\int\left(\frac{-3}{3 t+2}+\frac{2}{2 t+1}\right) d t$
$=\frac{-3 \log |3 t+2|}{3}+2 \frac{\log |2 t+1|}{2}+C$
$=-\log |3 t+2|+\log |2 t+1|+C$
$=\log \left|\frac{2 t+1}{3 t+2}\right|+C$
$=\log \left|\frac{2 \log x+1}{3 \log x+2}\right|+C \quad[\because t=\log x]$
Hence, the correct answer is option: 'B'.