What's the integral of ln(x)^2*x^2?

The integral of ln(x)^2*x^2 is (x^2/2)*(ln(x)^2-2ln(x)+2)+C.

To solve this integral, we can use integration by parts. Let u = ln(x)^2 and dv = x^2 dx. Then du/dx = 2ln(x)/x and v = (1/3)x^3. Using the formula for integration by parts, we have:

∫ln(x)^2*x^2 dx = u*v - ∫v*du/dx dx
= ln(x)^2*(1/3)x^3 - ∫(1/3)x^3*2ln(x)/x dx
= (x^2/3)*ln(x)^2 - (2/3)∫x^2ln(x) dx

To solve the remaining integral, we can use integration by parts again. Let u = ln(x) and dv = x^2 dx. Then du/dx = 1/x and v = (1/3)x^3. Using the formula for integration by parts, we have:

∫x^2ln(x) dx = u*v - ∫v*du/dx dx
= ln(x)*(1/3)x^3 - ∫(1/3)x^2 dx
= ln(x)*(1/3)x^3 - (1/9)x^3 + C

Substituting this back into the original equation, we get:

∫ln(x)^2*x^2 dx = (x^2/3)*ln(x)^2 - (2/3)*(ln(x)*(1/3)x^3 - (1/9)x^3) + C
= (x^2/2)*(ln(x)^2-2ln(x)+2)+C

Therefore, the integral of ln(x)^2*x^2 is (x^2/2)*(ln(x)^2-2ln(x)+2)+C.