Question

用换元法和分部积分法解积分∫x (lnx)^2 dx？

Answer 1

∫ x(lnx)² dx=∫ (lnx)² d(x²/2)
令u=(lnx)² ,v=x²/2,则
du = 2lnx * (1/x) dx
由分部积分公式∫u dv = uv - ∫v du
∫ x(lnx)² dx=∫ (lnx)² d(x²/2)
=(x²/2)(lnx)² - ∫(x²/2) * 2lnx * (1/x) dx
=(x²/2)(lnx)² - ∫x lnx dx
=(x²/2)(lnx)² - ∫ lnx d(x²/2)
=(x²/2)(lnx)² - [(x²/2) * lnx - ∫(x²/2) * (1/x) dx]
=(x²/2)(lnx)² - [(x²/2) * lnx - ∫(x/2) dx]
=(x²/2)(lnx)² - [(x²/2) * lnx - x²/4 ) + C
=(x²/4)*[2(lnx)²-2lnx+1]+C,2,令a=lnx
x=e^a
dx=e^ada
原式=∫e^a*a*e^ada
=∫ae^2ada
=1/2∫ae^2ad2a
=1/2∫ade^2a
=1/2ae^2a-1/2∫e^2ada
=1/2ae^2a-1/4e^2a+C
=1/2*lnx*x²-x²/4+C正确答案是(x^2/4)*[2*(lnx...,1,