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The primitive of 1(xa)3/2(bx)1/2 is

 

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a
1b−ab−xx−a1/2+C
b
34(b−a)b−xx−a1/2+C
c
1b−ax−ab−a1/2+C
d
none of these

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detailed solution

Correct option is D

putting x−a=(1/t) we have dx=−1/t2dt and the given integral can be written as ∫−1t2dt(1/t)3/2b−a−1t1/2=−∫dt(b−a)t−1=−2(b−a)t−1b−a+C=−2b−ab−a−(x−a)x−a+C=2a−bb−xx−a+C


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