高数问题 求人解答求人解答

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高数问题 求人解答求人解答
高数问题 求人解答

求人解答

高数问题 求人解答求人解答
y' = ∫dx/(1+x^2)
设 x = tant,则 dx = (sect)^2 *dt,t =arctanx
y'=∫(sect)^2 *dt/(sect)^2
  =∫dt
  = t + C1
  =arctanx + C1
y = ∫y' *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

(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