高等数学，请教这道不定积分怎么算? 高等数学 不定积分 这道题怎么算出来的？
\u8fd9\u9053\u9ad8\u6570\u6c42\u4e0d\u5b9a\u79ef\u5206\u600e\u4e48\u7b97\uff1f\u5206\u90e8\u79ef\u5206
\u222be^xsinxdx
=\u222bsinxde^x
=sinx*e^x-\u222be^xdsinx
=sinx*e^x-\u222be^xcosxdx
=sinx*e^x-\u222bcosxde^x
=sinx*e^x-cosx*e^x+\u222be^xdcosx
=sinx*e^x-cosx*e^x-\u222be^xsinxdx
\u79fb\u9879
\u6240\u4ee52\u222be^xsinxdx=sinx*e^x-cosx*e^x
\u6240\u4ee5\u222be^xsinxdx=e^x(sinx-cosx)/2
\u518d\u628a\u4e0a\u4e0b\u9650(0\u2192\u03c0/2)\u4ee3\u5165
=e^(\u03c0/2)/2 - (0-1)/2
=(1/2)[e^(\u03c0/2) +1]
=1/2·∫1/(x²+1)·d(x²+1)+arctanx
=1/2·ln(x²+1)+arctanx+C
【附注】
∫1/(x²+1)dx=arctanx+C
∫x/(x²+1)dx=1/2·ln(x²+1)+C
原式 = ∫ xdx/(1+x^2) + ∫ dx/(1+x^2)
= (1/2)∫ d(1+x^2)/(1+x^2) + ∫ dx/(1+x^2)
= (1/2)ln(1+x^2) + arctanx + C
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