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As why is in the for y=uv where u and v are funtions of x, dy/dx=u'v+v'u (where ' implies the derivative) u=(3x+1)², v=cos(3x) therefore using the chain rule u'=23(3x+1)=18x+6 and v'=-...

written out

Use integration by parts

let u=ln(x)

let dv/dx=x

therefore du/dx=1/x and v=(1/2)x^2

therefore the integral of xln(x) is equal to the following:

(1/2)x^2ln(x) - (integral...

36x^2 + 4

We'll first compute these intersections by setting x=0 and y=0 consecutively. This gives y=a-1 and a/(x-1)^2-1=0. Hence we find (x-1)^2=a, so x=1+-sqrt(a). As we have a>1 and we want the intersection w...