∫﹙㏑x﹚²dx
=(㏑x)²*x-∫x*2㏑x*(1/x)dx
=x(㏑x)²-2∫㏑xdx
=x(㏑x)²-2[x㏑x-∫x*(1/x)dx]
=x(㏑x)²-2x㏑x+2∫dx
=x(㏑x)²-2x㏑x+2x+C
∫cos﹙㏑x﹚dx
=∫cos(㏑x)dx
=x*cos(㏑x)-∫x*[-sin(㏑x)]*(1/x)dx
=xcos(㏑x)+∫sin(㏑x)dx
=xcos(㏑x)+[x*sin(㏑x)-∫xcos(㏑x)*(1/x)dx]
=xcos(㏑x)+xsin(㏑x)-∫cos(㏑x)dx
移项合并
=(x/2)[cos(㏑x)+sin(㏑x)]谢谢，我知道怎么写了∫㏑[x+√(1+x²)]dx =x*㏑[x+√(1+x²)]-∫x{1/[x+√(1+x²)]}*[1+x/√(1+x²)]dx=x㏑[x+√(1+x²)]-∫{x/[x+√(1+x²)]}*{[√(1+x²)+x]/√(1+x²)}dx=x㏑[x+√(1+x²)]-∫{x/√(1+x²)}dx=x㏑[x+√(1+x²)]-(1/2)∫{1/√(1+x²)}d(1+x²)=x㏑[x+√(1+x²)]-√(1+x²)+C