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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...

We know that non-real roots appear in complex conjugate pairs. Hence when we know one root, we know both of them. Then, as we can factorise a quadratic in it's linear factors, we know our quadratic is a c...