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\frac{1}{x^2+x}=\frac{A}{x} + \frac{B}{x+1}, 1=A(x+1)+Bx, let x=-1: 1=-B, B=-1let x=0: 1=A, A=1Hence, \int_{1}^{2}{\frac{1}{x^2+x}dx} = \int_{1}^{2}{\frac{1}{x}dx} - \int_{1}^{2}{\frac{1}{x+1}dx}=[ln(x)-l...

Using the product rule :u = x, so du/dx = 1
v = sin(x), so dv/dx = cos(x)
Therefore dy/dx = v(du/dx) + u(dv/dx) So dy/dx = sin(x) + x.cos(x)

We start by trying to find out the values of a and c using the information about the minimum point. We know we can rearrange the right hand side by completing the square: y = a(x^2 + 3x)+cy = a(x+3/2a)^2+...

When differentiating an expression of x and y we can use implicit differentiation when terms contain the y variable, recall that this is essentially differentiating the y term by y and then multiplying by...

x²>3(x+6)Firsty expand brackets --> x²>3x+18Then make one side = to 0 ---> x²-3x-18>0Solve the quadratic (x+3)(x-6)>0find the solutions when =0 x=-3 and ...