$\int \frac{(\log x-1)^2}{\left[1+(\log x)^2\right]^2} d x=$ (where $C$ is constant of integration.)

$\int \frac{(\log x-1)^2}{\left[1+(\log x)^2\right]^2} d x=$ (where $C$ is constant of integration.)
  1. $\frac{\log x}{(1+\log x)^2}+C$
  2. $\frac{e^{\log x}}{1+\log x}+C$
  3. $\frac{x}{1+(\log x)^2}+C$
  4. $\frac{\log x}{1+(\log x)^2}+C$

Solution

$\begin{aligned} & \int \frac{(\log x-1)^2}{\left\{1+(\log x)^2\right\}^2} \mathrm{~d} x=\int \frac{(\log x)^2+1-2 \log x}{\left\{1+(\log x)^2\right\}^2} \mathrm{~d} x \\ & =\int\left\{\frac{1}{1+(\log x)^2}-\frac{2 \log x}{\left\{1+(\log x)^2\right\}^2}\right\} \mathrm{d} x \\ & \int\left\{x\left\{\frac{-2 \log x}{x \cdot\left(1+(\log x)^2\right)^2}\right\}+\frac{1}{1+(\log x)^2}\right\} \mathrm{d} x \\ & =\frac{x}{1+(\log x)^2}+C\end{aligned}$

Asked in: MHT CET 2022 (11 Aug Shift 1)

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