MHT CET · Maths · Indefinite Integration
Let \(I=\int \tan ^{-1}\left(\frac{2 x}{1-x^2}\right) d x\), then \(I-2 x \tan ^{-1} x=\)
- A \(\log \left(1+x^2\right)+\mathrm{c}\)
- B \(-\log \left(1+x^2\right)+\mathrm{c}\)
- C \(-\log \left(1-x^2\right)+c\)
- D \(\log \left|\frac{2 \mathrm{x}}{1-\mathrm{x}^2}\right|+\mathrm{c}\)
Answer & Solution
Correct Answer
(B) \(-\log \left(1+x^2\right)+\mathrm{c}\)
Step-by-step Solution
Detailed explanation
\( \tan ^{-1}\left(\frac{2 x}{1-x^2}\right) = 2 \tan ^{-1} x \) \( I = \int 2 \tan ^{-1} x \, dx \)
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