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You're going to have to integrate by parts. Set

u=ln^2(x) and dv=1.

du/dx = 2*ln(x)*1/x and v=x.

u*v = ln^2(x)*x and v*du = 2*ln(x). So by the formula for integration by parts:

int(ln^2(x)) = ln^2(x)*x - int(2*ln(x) dx)

You can solve that second integral by parts as well, but you may already know that the integral of ln(x) is x*ln(x)-x. Substituting that in:

int(ln^2(x)) = ln^2(x)*x - 2*x*ln(x) + 2x

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The integral may be written as : 1/ln(2) ∫ 1/x dx = ln(x) /ln(2) + C

Integral ( [ln(x)]^2 dx )

To solve this, use integration by parts.

Let u = [ln(x)]^2. dv = dx

du = 2ln(x) (1/x) dx. v = x.

x[ln(x)]^2 - Integral ( 2x ln(x) dx )

Use parts one more time.

Let u = ln(x). dv = 2x dx

du = (1/x) dx. v = x^2

x[ln(x)]^2 - [ (1/x)x^2 - Integral ( (1/x)x^2 dx ) ]

x[ln(x)]^2 - [ x - Integral ( x dx ) ]

x[ln(x)]^2 - x + Integral (x dx )

x[ln(x]^2 - x + (1/2)x^2 + C

change of variable:

t = ln(x) => x = e^t

=> dt = dx/x

=> dx = x*dt = e^t * dt

Integrate ln^2(x) dx = Integrate (e^t) (t^2) dt

Now take the integral by part...(it's elementary)

Ok, let's solve it :

integrate((lnx)^2*dx)

x = e^t >>>> dx = e^t

integrate(t^2*e^t)*dt

integrating part by part :

t^2 = u >>>> du = 2t*dt

e^t = dv >>>> v = e^t

u*v - integrate(v*du)

t^2*e^t - 2integrate(e^t*tdt) .....(*)

Now, let's integrate : integrate(e^t*t*dt)

u = t >>>>>> du = dt

e^t = dv >>>> v = e^t

e^t*t - integrate(e^t*dt) = e^t*t - e^t

replacing of (*)

t^2*e^t - 2integrate(e^t*tdt)

t^2*e^t - 2e^t*t + 2e^t

2e^t + t^2*e^t - 2e^t*t

remember : x = e^t

2x + (lnx)^2*x - 2x*lnx

Hope that helps

so glad that I found this topic already answered! It is like you've read my mind!

sin^2(x)

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