MHT CET · Maths · Indefinite Integration
\(\int \frac{(\log x-1)^2}{\left[1+(\log x)^2\right]^2} d x=\) (where \(C\) is constant of integration.)
- A \(\frac{\log x}{(1+\log x)^2}+C\)
- B \(\frac{e^{\log x}}{1+\log x}+C\)
- C \(\frac{x}{1+(\log x)^2}+C\)
- D \(\frac{\log x}{1+(\log x)^2}+C\)
Answer & Solution
Correct Answer
(C) \(\frac{x}{1+(\log x)^2}+C\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \int \frac{(\log x-1)^2}{\left\{1+(\log x)^2\right\}^2} \mathrm{~d} x=\int \frac{(\log x)^2+1-2 \log x}{\left\{1+(\log x)^2\right\}^2} \mathrm{~d} x \\ & =\int\left\{\frac{1}{1+(\log x)^2}-\frac{2 \log x}{\left\{1+(\log x)^2\right\}^2}\right\} \mathrm{d} x \\ & \int\left\{x\left\{\frac{-2 \log x}{x \cdot\left(1+(\log x)^2\right)^2}\right\}+\frac{1}{1+(\log x)^2}\right\} \mathrm{d} x \\ & =\frac{x}{1+(\log x)^2}+C\end{aligned}\)
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