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B
43(x−2x+1)1/4+c
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C
13(x−1x+2)1/4+c
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D
13(x+2x−2)1/4+c
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Solution
The correct option is A43(x−1x+2)1/4+c ∫1[(x−1)3(x+2)5]1/4dx=∫1(x−1x+2)34(x+2)2dx=13∫1t34dt[∵x−1x+2=t⇒3(x+2)2dx=dt]=13(t1/414)+c=43tt4+c=43(x−1x+2)1/4+c