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Step 1: multiply (5sqrt(3)-6)/(2sqrt(3)+3) by (2sqrt(3)-3)/(2sqrt(3)-3) (this is a trick called CONJUGATION as you're really just multiplying the fraction by 1. Step 2: Expand and ...

factorise the top fraction: 3x^2 - x - 2 = (3x + 2)(x - 1)

factorise the bottom fraction: x^2 -1 = (x + 1)(x - 1)

cancel out the (x - 1) and the final result is (3x + 2) / (x + 1)

to find the area integrate the function between 0 and pi/2:

indefinite integral of y = -1/2 cos(2x) + sin(x) + c

ignore c and input boundary conditions:

[-1/2 cos(pi) + sin(pi/2)] - [...

rewrite as y = 4x^5 - 5x^(-2)

dy/dx = 20x^4 + 10x^(-3)

integral of y = 4x^6 / 6 + 5x^(-1) / 1 + c

simplify: 2/3 x^6 + 5/x

i) There first step is to acknowledge the need for both product rule (d(uv)/dx=v.du/dx+u.dv/dx) and chain rule (dz/dx=dz/dy*dy/dx). Here, u=x^3 and v=ln(2x). Therefore, du/dx=3x^2 which is a standard diff...