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Question Number 104388 by Ar Brandon last updated on 21/Jul/20
∫dxx2x3−13
Answered by Dwaipayan Shikari last updated on 21/Jul/20
∫dxx3(1−1x3)13=∫1x4.x(1−1x3)13dx=x3∫3x4(1−1x3)13−13∫∫3x4(1−1x3)13=x2(1−1x3)23−12∫(1−1x3)23=x2(1−1x3)23−12∫3x4.x43(1−1x3)23dxx2(1−1x3)23−16∫x4t23dt{1−1x3=tx3=11−tx2(1−1x3)−16∫(11−t)43t23.....continue
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