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JEE Mains · Maths · STD 12 - 7.1 indefinite integral
If \(\int {\frac{{{x^2} - x + 1}}{{{x^2} + 1}}{e^{{{\cot }^{ - 1}}}}{}^x\,dx = A(x)} {e^{{{\cot }^{ - 1}}}}{}^x + C,\) then \(A(x)\) is equal to
- A \(-x\)
- B \(x\)
- C \(\sqrt {1-x}\)
- D \(\sqrt {1+x}\)
Answer & Solution
Correct Answer
(B) \(x\)
Step-by-step Solution
Detailed explanation
Let \(I = \int {\frac{{{x^2} - x + 1}}{{{x^2} + 1}}} \cdot {e^{{{\cot }^{ - 1}}x}}dx\) Put \(x = \cot t\) \( \Rightarrow - \cos e{c^2}tdt = dx\) Now, \(1 + {\cot ^2}t = \cos e{c^2}t\) \(\therefore \)…
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