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y&#39; = ∫dx/(1+x^2)设 x = tant,则 dx = (sect)^2 *dt,t =arctanxy&#39;=∫(sect)^2 *dt/(sect)^2& =∫dt& = t + C1& =arctanx + C1y = ∫y&#39; *dx&& = ∫(arctanx + C1)*dx&& = ∫arctanx *dx + C1*∫dx&& = x*arctanx - ∫x*dx/(1+x^2) + C1*x&& = x*arctanx - 1/2*∫d(1+x^2)/(1+x^2) + C1*x&& = x*arctanx - 1/2*ln(1+x^2) + C1*x + C∫x*(x^2 + 1)^2 *dx=1/2*∫(x^2+1)^2 *(2x*dx)=1/2*∫(x^2+1)^2 *d(x^2 +1)=1/2*1/3*(x^2+1)^3 + C=1/6*(x^2+1)^3 + C
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（5）y''=1/（1+x^2）令y'=dy/dx=p 则y''=dp/dx所以有dp/dx=1/（1+x^2）积分得 p=arctanx+C1即 dy/dx=arctanx+C1dy=(arctanx+C1)dx两边积分有 y=∫ (arctanx+C1)dx=∫ arctanxdx+C1x=xarctanx-∫ x/(1+x^2)dx+C1x=xarctanx-1/2∫ 1/(1+x^2)d(1+x^2)+C1x=xarctanx-1/2ln(1+x^2)+C1x+C2
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