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Question Number 122798 by aurpeyz last updated on 19/Nov/20
Commented by mohammad17 last updated on 19/Nov/20
I=∫131x2(1+3x)23dx=−15(2)53+15(4)53
Answered by MJS_new last updated on 19/Nov/20
∫1x2(x+3x)2/3dx=[t=(x+3x)1/3→dx=−(x+3)2/3x4/3dt]=−∫t4dt=−t55=−(x+3)5/35x5/3+C⇏∫311x2(x+3x)2/3dx=2235(4−23)
Answered by Dwaipayan Shikari last updated on 20/Nov/20
∫131x2(x+3x)23dx=−16∫13−6x2(1+3x)23dx=16∫24t23dt=110[t53]24=4523−15223
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