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B
x⋅ecot−1x
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C
ecot−1x1+x2
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D
ecot−1x
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Solution
The correct option is Bx⋅ecot−1x ∫ecot−1x(x2+1−x1+x2)dx=∫ecot−1x(1−x1+x2)dx =∫ecot−1xdx−∫ecot−1x(x1+x2)dx =ecot−1x⋅x−∫−ecot−1x(11+x2)⋅xdx−∫ecot−1x(x1+x2)dx =ecot−1x⋅x+C