求sinx/(asinx+bcosx)的不定积分 x^2/(x^2+2x+2)^2的不定积分

Let sinx = R(asinx+bcosx) + S(acosx-bsinx) + T = (Rb+Sa)cosx + (Ra-Sb)sinx + T
Rb+Sa=0,Ra-Sb=1,T=0
S=-Rb/a
Ra-(-Rb/a)b=1,R=a/(a²+b²)
S=-b/(a²+b²)
I = ∫ sinx/(asinx+bcosx) dx
= [a/(a²+b²)]∫ (asinx+bcosx)/(asinx+bcosx) dx + [-b/(a²+b²)]∫ (acosx-bsinx)/(asinx+bcosx) dx
= [a/(a²+b²)]∫ dx - [b/(a²+b²)]∫ d(asinx+bcosx)/(asinx+bcosx)
= [a/(a²+b²)]x - [b/(a²+b²)]ln| asinx+bcosx | + C
∫ x²/(x²+2x+2) dx
= ∫ [(x²+2x+2)-(2x+2)]/(x²+2x+2) dx
= ∫ (x²+2x+2)/(x²+2x+2) dx - ∫ (2x+2)/(x²+2x+2) dx
= ∫ dx - ∫ d(x²+2x+2)/(x²+2x+2)
= x - ln| x²+2x+2 | + C第二问你看错题了，少了个平方J = ∫ x^2/(x^2+2x+2) dx= ∫ x^2/[(x^2+2x+1)+1] dx= ∫ x^2/[(x+1)^2+1] dxLet y = x+1，dy = dxJ = ∫ (y-1)^2/(y^2+1) dy= ∫ [1/(y^2+1) - 2y/(y^2+1)^2] dy= ∫ dy/(y^2+1) - ∫ d(y^2+1)/(y2+1)^2= arctan(y) + 1/(y^2+1) + C= arctan(x+1) + 1/[(x+1)^2+1] + C= arctan(x+1) + 1/[(x^2+2x+2) + CNote：(y-1)^2/(y^2+1)^2 = (y^2-2x+1)/(y^2+1)^2 = (Ay+B)/(y^2+1) + (Cy+D)/(y^2+1)^2y^2-2y+1 = (Ay+B)(y^2+1) + Cy+Dy^2-2y+1 = Ay^3+By2+Ay+B+Cy+Dy^2-2y+1 = Ay^3+By^2+(A+C)y+(B+D)A=0B=1A+C=-2 => C=-2B+D=1 => D=1-1=0∴(y-1)^2/(y^2+1)^2 = 1/(y^2+1) - 2y/(y^2+1)^2