We have to find the integral of y= x / sqrt(x^2+9)

Int [ x/ sqrt (x^2 + 9) dx]

let x^2 + 9 = u, du / 2 = x dx

=> (1/2)*Int [ 1/ sqrt u du]

=> (1/2)*Int [ u^(-1/2) du]

=> (1/2) u^(1/2) / (1/2)

=> u^(1/2)

substitute u = x^2 + 9

=> sqrt (x^2 + 9) + C

**The required integral is sqrt (x^2 + 9) + C**

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