Question

求解两道不定积分题

1.ln平方xdx 2.x平方cos平方(x/2)dx

Answer 1

如图所示

Answer 2

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

2
∫.x^2cos^2(x/2)dx
∫.x^2(1+cosx)/2dx
=∫(1/2)x^2dx+∫(1/2)x^2*cosxdx
=1/6x^3+(1/2)∫x^2*cosxdx
∫x^2*cosxdx=x^2*sinx-∫2xsinxdx
=x^2*sinx+2xcosx-2∫cosxdx
=x^2*sinx+2xcosx-2sinx
原式=1/6x^3+(1/2)x^2*sinx+xcosx-sinx