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∫ u(dv/dx) dx = uv − ∫ v(du /dx)dx is the Integration by Parts formula. 

If you set u=lnx, differentiation (rememeber from tables) leads to du/dx= 1/x, and dv/dx=x and so v=x^2/2 (raise power by on...

This question is aimed at A-Level Pure Core 1 students. A circle with centre C(2, 3) passes through the point A(-4,-5).  Find the equation of the circle in the form (x-a)^2 + (y-b)^2 = k ...

To differentiate, we bring the power down and decrease the power by 1. So x³ becomes 3x², -3x² becomes -6x, and -9x (which can be written as -9x¹ ) becomes -9. ...

This type of question appears over-complicated with limited options, but one must not fear! At first, the numerator seems to be of a higher degree than the denominator (x^4 compared to x^2), but from the ...

Using the product rule, f=uv, df = (vu'-uv')/v^2. we first set u = 3x^2 and v = sin(2x). u' = 6x, v'=2cos(2x) Therefore, vu' = 6xsin(2x). uv' = 6x^2cos(2x), v^2 = 4cos^2(2x) Therefore the differe...