ok, so yes, for people like us who dont use LaTex, x^2 means x squared. personally i use sqrt() to mean squareroot, and int{}dx for integrate something with respect to x

1. Find int{3x+ 5sqrt(x)}dx

int{3x+ 5sqrt(x)}dx

= int{3x+ 5x^(1/2)}dx

= (3/2)x^2 + (10/3)x^(3/2) + C

remember, to integrate a function of the form x^n, it is int{x^n}dx = (x^(n+1))/n+1. as in, you add 1 to the power and divide by the new power, this is called the "power rule"

2. find int{(X-1)(2X+5)}dx

int{(X-1)(2X+5)}dx

= int{2x^2 + 3x - 5}dx .............expand the brackets

= (2/3)x^3 + (3/2)x^2 - 5x + C

3. find f 1/2X^2 - 3X^1/2 - 5X^-1/3 dx

is it (1/2)x^2 or 1/(2x^2)?

4. find f (X+2)^2 / X^1/2 dx

int{[(x+2)^2]/x^(1/2)}dx

= int{(x^2 + 4x + 4)/x^(1/2)}dx

= int{x^(3/2) + 4x^(1/2) + 4x^(-1/2)}dx .............do you understand this step? i divided each term of the top by x^(1/2)

= (2/5)x^(5/2) + (8/3)x^(3/2) + 8x^(1/2) + C

5. given that dy/dx = 2X+X^-2 and that y=2 at x=1 find y in terms of X then the value of y at X=-1

dy/dx means the derivative of y with respect to x. this means the function you see was differentiated, we have to integrate to get back to the original function.

dy/dx = 2x + x^-2

=> dy = (2x + x^-2)dx

=>int{1}dy = int{2x + x^-2}dx

=> y = x^2 - x^-1 + C

when y=2, x = 1

=> 2 = 1^2 - 1^-1 + C

=> 2 = C

so our original function is y = x^2 - 1/x + 2

now to find y at x = -1

y = (-1)^2 - 1/-1 + 2

y = 1 + 1 + 2 = 4