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JEE Mains · Maths · STD 12 - 7.2 definite integral

If  \(2\int_0^1 {{{\tan }^{ - 1}}}\,xdx = \int_0^1 {{{\cot }^{ - 1}}}\,(1 - x + {x^2})dx,\) then \(\int_0^1 {{{\tan }^{ - 1}}}\, (1 - x + {x^2})dx\)  is equal to 

  1. A \(\frac{\pi }{2} + \log \,2\)
  2. B \(\log \,2\)
  3. C \(\frac{\pi }{2} - \log \,4\)
  4. D \(\log \,4\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\log \,2\)

Step-by-step Solution

Detailed explanation

\(2 \int_{0}^{1} \tan ^{-1} x d x=\int_{0}^{1}\left(\frac{\pi}{2}-\tan ^{-1}\left(1-x+x^{2}\right)\right) d x\) \(2 \int_{0}^{1} \tan ^{-1} x d x=\int_{0}^{1} \frac{\pi}{2} d x-\int_{0}^{1} \tan ^{-1}\left(1-x+x^{2}\right) d x\)…
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