int2xlnx^2dx?

Let u = x²
Then, du = 2x dx

∫2x ln(x²) dx
= ∫ln(x²) (2x dx)
= ∫ln(u) du
= u ln(u) - u + C
= x² ln(x²) - x² + C
= 2x² ln(x) - x² + C

∫ 2x ln x^2 dx

Let t=x^2
dt = 2x dx

∫ 2x ln x^2 dx = ∫ ln t dt

Integrate ∫ ln t dt by parts
dv= dt; v=t
u = ln t ; du = 1/t dt

∫ u dv = uv - ∫ v du
∫ ln t dt = (ln t) t - ∫ t (1/t) dt
∫ ln t dt = (ln t) t - ∫ dt
∫ ln t dt = (ln t) t - t

= t ln t - t

replace t by x^2
= x^2 ln x^2 - x^2 + C

No