MHT CET · Maths · Definite Integration
The value of \(\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x\) is
- A 1
- B 0
- C \(-1\)
- D None of these
Answer & Solution
Correct Answer
(B) 0
Step-by-step Solution
Detailed explanation
Let
\(\begin{aligned} I &=\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x \\ &=\int_{0}^{1} \tan ^{-1}\left(\frac{x+x-1}{1-x(x-1)}\right) d x \\ &=\int_{0}^{1} \tan ^{-1} x d x+\int_{0}^{1} \tan ^{-1}(x-1) d x \\ &=\int_{0}^{1} \tan ^{-1} x d x+\int_{0}^{1} \tan ^{-1}(1-x-1) d x \\ &=\int_{0}^{1} \tan ^{-1} x d x-\int_{0}^{1} \tan ^{-1} x d x=0 \end{aligned}\)
\(\begin{aligned} I &=\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x \\ &=\int_{0}^{1} \tan ^{-1}\left(\frac{x+x-1}{1-x(x-1)}\right) d x \\ &=\int_{0}^{1} \tan ^{-1} x d x+\int_{0}^{1} \tan ^{-1}(x-1) d x \\ &=\int_{0}^{1} \tan ^{-1} x d x+\int_{0}^{1} \tan ^{-1}(1-x-1) d x \\ &=\int_{0}^{1} \tan ^{-1} x d x-\int_{0}^{1} \tan ^{-1} x d x=0 \end{aligned}\)
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