Evalaute `intx^(_1//2)(2+3x^(1//2))^(-2)dx`.

Explanation

`I=intx^(-1//2)(2+3x^(1//3))^(-2)dx`
Let ` x=t^(6) " or " dx=6t^(5)dt`
` :. I=int t^(-3)(2+3t^(2))^(-2) *6t^(5) dt`
`=6 int (t^(2))/((2+3t^(2))^(2))dt`
`=(6)/(9)int (t ^(2)dt)/(((2)/(3)+t^(2))^(2))`
Now, Let ` t=sqrt(((2)/(3)))tan theta`
` :. dt =sqrt(((2)/(3)))sec^(2) theta d theta`
` :. I=(6)/(9)int((2)/(3)tan^(2) theta *sqrt(((2)/(3)))sec^(2) theta d theta)/((4)/(9)sec^(4) theta)`
`=sqrt((2)/(3))intsin^(2) theta d theta`
`=(1)/(sqrt(6))int(1-cos 2 theta)d theta`
`=(1)/(sqrt(6)){theta - (sin 2 theta)/(2)}+c`
`=(1)/(sqrt(6)){theta - (tan theta)/(1+tan^(2)theta)}+c`
`=(1)/(sqrt(6)){"tan"^(-1){sqrt((3)/(2))t}-(sqrt((3)/(2))*t)/(1+(3)/(2)t^(2))}+c`
`=(1)/(sqrt(6)){"tan"^(-1){sqrt((3)/(2))x^(1//6)}-(sqrt(6)x^(1//6))/(2+3x^(1//3))}+c`