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The value of the integral int3^6 sqrtx/(...

The value of the integral `int_3^6 sqrtx/(sqrt(9-x)+sqrtx)dx` is

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The correct Answer is:
`3//2`
`I=int_(3)^(6) (sqrt(x))/(sqrt(9-x)+sqrt(x)) dx`………….1
`:. I=int_(3)^(6)(sqrt(9-x))/(sqrt(x)+sqrt(9-x))dx`…………2
Adding 1 and 2.
`2I=int_(3)^(6)1.dx=[x]_(3)^(6)=6-3=3`
Hence `I=3/2`
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