∫1/﹙1+x²﹚²dx=?

显示全部楼层 · 2013-1-23 21:42:41

令x = tanθ，dx = sec2θ dθ= ∫ 1/(1 + x2)2 dx= ∫ 1/(1 + tan2θ)2 ? (sec2θ dθ)= ∫ 1/sec?θ ? sec2θ dθ= ∫ cos2θ dθ= (1/2)∫ (1 + cos2θ) dθ= (1/2)(θ + 1/2 ? sin2θ) + C= θ/2 + (1/2)sinθcosθ + C= (arctanx)/2 + (1/2)[x/√(x2 + 1)][1/√(x2 + 1)] + C= (ar...

千问 · 2013-1-23 21:42:41

因为∫1/(1+x^2)^ndx=x/(1+x^2)^n-∫x*(-n)/(1+x^2)^(n+1)*2xdx=x/(1+x^2)^n+n∫(2x^2+2-2)/(1+x^2)^(n+1)dx=x/(1+x^2)^n+2n∫1/(1+x^2)^ndx-2n∫1/(1+x^2)^(n+1)dx所以∫1/(1+x^2)^(n+1)dx=x/(2n(1+x^2...