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B
320(1+x2)23(2x2−3)+C
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C
320(1+x2)23(2x2+3)+C
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D
None of these
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Solution
The correct option is B320(1+x2)23(2x2−3)+C Put 1+x2=t3⇒2xdx=3t2dt∴∫x3(1+x2)1/3dx=∫x2.x(1+x2)1/3dx=32∫(t3−1)tdt=32∫(t4−t)dt=32(t55−t22)+C=320t2(2t3−5)+C=320(1+x2)23(2x2−3)+C