How to Integrate ∫ln^2(x)dx

In this video, we work through how to integrate ln^2(x). This integral follows the same approach for the integral of ln(x), and hence the method we use is integration by parts.
Integration by parts is written as...
∫udv = uv - ∫vdu
In this case we let:
u = ln^2(x) and...
dv = dx
Please watch the video for the full tutorial
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Integral ln^2 (sin x) pls explain

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Hi sir please can you help me
I want to know how to solve this
Sketch the graph of the function:
Y=sin2x and y=1-sinx

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[@gelbkehlchen]

Solution:
∫ln²(x)*dx =
-----------------------------------------
Substitution: u = ln(x) x = e^u du = 1/x*dx dx = x*du = e^u*du
-----------------------------------------
= ∫u²*e^u*du =
------------------------------------
Solution by partial integration:
Partial integration can be derived from the product rule of differential calculus. The product rule of differential calculus states:
(u*v)’ = u’*v+u*v’ |-u’*v ⟹
(u*v)’-u’*v = u*v’ |∫() ⟹
∫u*v’*dx = u*v-∫u’*v*dx
------------------------------------
= u²*e^u-2*∫u*e^u*du = u²*e^u-2*(u*e^u-∫e^u*du) = u²*e^u-2*(u*e^u-e^u)+C
= u²*e^u-2*u*e^u+2*e^u+C
= ln²(x)*x-2*ln(x)*x+2*x+C
= [ln²(x)-2*ln(x)+2]*x+C
Check the result by deriving:
{[ln²(x)-2*ln(x)+2]*x+C}’ = [2*ln(x)/x-2/x]*x+ln²(x)-2*ln(x)+2
= 2*ln(x)-2+ln²(x)-2*ln(x)+2 = ln²(x) everything o.k.

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