MHT CET · Maths · Definite Integration
\(\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x=\)
- A 1
- B 4
- C 2
- D 0
Answer & Solution
Correct Answer
(D) 0
Step-by-step Solution
Detailed explanation
(A)
\(\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x\)
\(=\int_{0}^{1} \tan ^{-1}\left[\frac{x-(1-x)}{1+x(1-x)}\right] d x=\int_{0}^{1}[\tan ^{-1} x-\tan ^{-1}\) \((1-x)]\) \(d x\)...(1)
\(=\int_{0}^{1}\left[\tan ^{-1}(1-x)-\tan ^{-1}(1-(1-x))\right] d x=\int_{0}^{1}\) \(\left[\tan ^{-1}(1-x)-\tan ^{-1}(x)\right] d x\)...(2)
Adding (1) \& (2), we get
\(2 I=0 \Rightarrow I=0\)
\(\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x\)
\(=\int_{0}^{1} \tan ^{-1}\left[\frac{x-(1-x)}{1+x(1-x)}\right] d x=\int_{0}^{1}[\tan ^{-1} x-\tan ^{-1}\) \((1-x)]\) \(d x\)...(1)
\(=\int_{0}^{1}\left[\tan ^{-1}(1-x)-\tan ^{-1}(1-(1-x))\right] d x=\int_{0}^{1}\) \(\left[\tan ^{-1}(1-x)-\tan ^{-1}(x)\right] d x\)...(2)
Adding (1) \& (2), we get
\(2 I=0 \Rightarrow I=0\)
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