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Please see this document which contains a scanned copy of my answer.https://docs.google.com/document/d/1A9idRZsUAy5QMLSJNDQEun-l7mSEjpRLx_Jwx44JoqY/edit?usp=sharing

To solve this inequality, we need to take steps to collect like terms on each side of the inequality. To start with, we can try getting all the numbers on one side of the inequality - on the left hand sid...

You need to remember some ‘rules’:if it has the ‘co-‘ prefix, the derivative will be negative (e.g. cos x -> -sin x)if it has the ‘-sec’ suffix, the derivative will only include itself (e.g. cosec x -&...

\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)