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Question Number 120582 by MagdiRagheb last updated on 01/Nov/20
Letu=x35du=35x25I=53∫u3−2udu=−56∫3−2u−33−2udu=−53∫[3−2u−3(3−2u)−12]du=−53[−13(3−2u)32+3(3−2u)12]+c=−518(3−2u)12[−3+2u+9]+c=−518(3−2u)12(6+2u)+c=−593−2x35(3+x35)+c
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![Let u=x^(3/5) du = (3/(5x^(2/5) )) I = (5/3)∫(u/( (√(3−2u)))) du = −(5/6) ∫((3−2u−3)/( (√(3−2u)))) du = −(5/3) ∫[(√(3−2u)) −3(3−2u)^(−(1/2)) ] du = −(5/3)[−(1/3)(3−2u)^(3/2) +3(3−2u)^(1/2) ]+c = −(5/(18))(3−2u)^(1/2) [−3+2u + 9]+c = −(5/(18))(3−2u)^(1/2) (6+2u)+c = −(5/9)(√(3−2x^(3/5) ))(3+x^(3/5) )+c](Q120582.png)