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Question Number 82019 by M±th+et£s last updated on 17/Feb/20

Answered by mind is power last updated on 18/Feb/20

2F1(k+1,k+1;k+2;1)=(k+1)∫_0 ^1 x^k (1−x)^(−k−1) dx  not defind sir tchek this Quation

2F1(k+1,k+1;k+2;1)=(k+1)01xk(1x)k1dxnotdefindsirtchekthisQuation

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