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Question Number 142945 by Rankut last updated on 07/Jun/21
∫00.51+x33dx
Answered by mindispower last updated on 07/Jun/21
(1+y)a=1+∑n⩾1∏n−1k=0(a−k)n!xn(1+x3)13=1+∑n⩾1∏n−1k=0(−1)n(k−13)n!xn=12(1+∑n⩾1(k−13)n.n!(n+1)!.(−12)nn!)=122F1(−13,1;2;[−12])
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