CUET · MATHS · PYQ PAPER 2025
\(\int_0^1 \tan ^{-1}\left(\frac{2 x-1}{1+x-x^2}\right) d x\) is equal to :
- A \(-1\)
- B 0
- C 1
- D \(\frac{\pi}{4}\)
Answer & Solution
Correct Answer
(B) 0
Step-by-step Solution
Detailed explanation
\(I = \int_0^1 \tan^{-1}\left(\frac{2x-1}{1+x-x^2}\right) dx\) Using \(\int_0^a f(x) dx = \int_0^a f(a-x) dx\): \(I = \int_0^1 \tan^{-1}\left(\frac{2(1-x)-1}{1+(1-x)-(1-x)^2}\right) dx\) \(I = \int_0^1 \tan^{-1}\left(\frac{2-2x-1}{1+1-x-(1-2x+x^2)}\right) dx\)…
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