(ln(x))^2 dx, integrate.
use integration by parts. That is just the product rule in reverse.
d(uv) = u dv + v du
u dv = d(uv) - v du
∫u dv = ∫d(uv) - ∫v du
∫u dv = uv - ∫v du
So, here we just let
u = (lnx)^2
dv = dx
du = 2lnx * 1/x dx = (2lnx)/x
v = x
∫(lnx)^2 dx = x(lnx)^2 - ∫2lnx dx
Now repeat, this time letting
u = lnx
dv = 2dx
To integrate the function (ln(x))^2 dx, we can use integration by parts. The formula for integration by parts is:
∫u * v dx = u * ∫v dx - ∫(u' * ∫v dx) dx
Let's break down the given function and assign u and dv accordingly:
u = (ln(x))^2 --> u' = 2ln(x)/x
dv = dx --> v = x
Now we can apply the integration by parts formula:
∫(ln(x))^2 dx = u * ∫v dx - ∫(u' * ∫v dx) dx
= (ln(x))^2 * ∫x dx - ∫(2ln(x)/x * ∫x dx) dx
= (ln(x))^2 * (x^2/2) - ∫(2ln(x)/x * x^2/2) dx
= (ln(x))^2 * (x^2/2) - ∫ln(x) * x dx
To integrate the last term, we can use integration by parts again:
u = ln(x) --> u' = 1/x
dv = x dx --> v = x^2/2
Applying the integration by parts formula once again:
∫ln(x) * x dx = u * ∫v dx - ∫(u' * ∫v dx) dx
= ln(x) * (x^2/2) - ∫(1/x * x^2/2) dx
= ln(x) * (x^2/2) - ∫x/2 dx
= ln(x) * (x^2/2) - (x^2/4) + C
Therefore, the result of the integral is:
∫(ln(x))^2 dx = (ln(x))^2 * (x^2/2) - (x^2/4) + C
where C is the constant of integration.