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∫1/x∧2Dx=

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

设x=sint, 那么dx=cost dt 而1/(1-x^2)^3/2 =1/ (cost)^3 所以得到 原积分=∫ 1/ (cost)^3 *cost dt =∫ 1/ (cost)^2 dt =tant +C =x /(1-x^2)^1/2 +C,C为常数

∫[(x-1)/(x²+2x+2)]dx =½∫[(2x+2-4)/(x²+2x+2)]dx =½∫[(2x+2)/(x²+2x+2)]dx -2∫d(x+1)/[1+(x+1)²] =½ln(x²+2x+2)-2arctan(x+1) +C

郭敦荣回答: 若换元时,则应换到底;若不换元,则积分上下限不变。 换元法,令u=1+x²,则du=d(1+x²)=2xdx。x=0时,u=1;x=1时,u=2, 于是,原式=(1/2)∫(1,2)(1/u)du=(1/2) lnu|(1,2) =(1/2)(ln2-ln1) =(1/2)ln...

1/x^2在x=0处不连续,结果不存在

令u=1+x^2 则du=2xdx 原式=1/2·∫1/udu =1/2·lnu+C =1/2·ln(1+x^2)+C

三角换元来做;有x^2和x^2+1,利用tan换元;过程如下:令x=tanu,则x²+1=sec²u,dx=sec²udu ∫x^2/(x^2+1)^2dx =∫ [tan²u/(secu)^4]sec²udu =∫ tan²u/sec²udu =∫ (sec²u-1)/sec²udu =∫ 1 du - ∫ cos&...

见图

∫x/(1+x^2)dx =1/2∫1/(1+x^2)dx^2 =1/2∫1/(1+t)dt =1/2ln|1+t|+C =1/2ln(1+x^2)+C

∫ (1-x-x²)/(x²+1)² dx =∫ (2-x-x²-1)/(x²+1)² dx =2∫ 1/(x²+1)² dx - ∫ x/(x²+1)² dx - ∫ (x²+1)/(x²+1)² dx =2∫ 1/(x²+1)² dx - (1/2)∫ 1/(x²+1)² dx&...

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