Q105 / 160 · AP EAMCET 2015 · Mathematics

$\int \frac{f(x) g^{\prime}(x)-f^{\prime}(x) g(x)}{f(x) g(x)} \times[\log g(x)-\log f(x)] d x$ is equal to

  • A
    $\log \left[\frac{g(x)}{f(x)}\right]+C$
  • Correct answer
    B
    $\frac{1}{2}\left[\log \frac{g(x)}{f(x)}\right]^2+C$
  • C
    $\frac{g(x)}{f(x)} \log \left[\frac{g(x)}{f(x)}\right]+C$
  • D
    $\log \left[\frac{g(x)}{f(x)}\right]-\frac{g(x)}{f(x)}+C$

Solution

Let $I=\int \frac{f(x) g^{\prime}(x)-f^{\prime}(x) g(x)}{f(x) g(x)}$ $\begin{aligned} & {[\log g(x)-\log f(x)] d x} \\ & =\int \log \left[\frac{g(x)}{f(x)}\right] \cdot d\left\{\log \frac{g(x)}{f(x)}\right\} \end{aligned}$ Put $t=\log \left[\frac{g(x)}{f(x)}\right]$ $d t=d\left(\log \left[\frac{g(x)}{f(x)}\right]\right)$ Now, $\quad I=\int t d t=\frac{t^2}{2}+C$ $=\frac{1}{2}\left\{\log \frac{g(x)}{f(x)}\right\}^2+C$

Hence, the correct answer is option: 'B'.

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