求不定积分 ∫﹙㏑x﹚²dx 和 ∫cos﹙㏑x﹚dx
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求不定积分 ∫﹙㏑x﹚²dx 和 ∫cos﹙㏑x﹚dx
求不定积分 ∫﹙㏑x﹚²dx 和 ∫cos﹙㏑x﹚dx,
求不定积分 ∫﹙㏑x﹚²dx 和 ∫cos﹙㏑x﹚dx,
![求不定积分 ∫﹙㏑x﹚²dx 和 ∫cos﹙㏑x﹚dx](/uploads/image/z/2172103-7-3.jpg?t=%E6%B1%82%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86+%E2%88%AB%EF%B9%99%E3%8F%91x%EF%B9%9A%26%23178%3Bdx+%E5%92%8C+%E2%88%ABcos%EF%B9%99%E3%8F%91x%EF%B9%9Adx)
∫﹙㏑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)]
再问: 谢谢,我知道怎么写了
再答: ∫㏑[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
=(㏑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)]
再问: 谢谢,我知道怎么写了
再答: ∫㏑[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