Bitsat
Indefinite Integration
BITSAT 2022
MCQ+3 / -12022
Let \(f(x) = \int {{{{x^2}dx} \over {(1 + {x^2})(1 + \sqrt {1 + {x^2}} )}}}\) and \(f(0) = 0\), then the value of \(f(1)\) be
Bitsat
BITSAT 2022
Let \(f(x) = \int {{{{x^2}dx} \over {(1 + {x^2})(1 + \sqrt {1 + {x^2}} )}}}\) and \(f(0) = 0\), then the value of \(f(1)\) be