w_noice
花朵
共回答了24个问题采纳率:87.5% 举报
(1) ∫[0,π](xsinx)^2dx
=1/2∫[0,π]x^2(1-cos2x)dx
=1/2∫[0,π]x^2dx-1/4∫[0,π]x^2dsin2x
=1/6x^3|[0,π]-1/4x^2 sin2x|[0,π]+1/2∫[0,π]xsin2xdx
=π^3/6-1/4∫[0,π]xdcos2x
=π^3/6-1/4xcos2x|[0,π]+1/4∫cos2xdx
=π^3/6-π/4+1/8sin2x|[0,π]
=π^3/6-π/4
(2) ∫[0,2]xdx/(x^2-2x+2)^2 (疑错,已改)
=∫[0,2][(x-1)+1]dx/(x^2-2x+2)^2
=-1/2*1/(x^2-2x+2)|[0,2]+∫[0,2]d(x-1)/[(x-1)^2+1]^2
=1/2*(x-1)/[(x-1)^2+1]|[0,2]+1/2arctan(x-1)|[0,2]
=1/4+1/4+π/8+π/8
=1/2+π/4
(3) F(x)=∫[0,x]sintdt/t
F'(x)=sinx/x
F'(0)=lim[x-->0]sinx/x=1
(4) ∫(3x^4+2x^2)dx/(x^2+1)
=∫[3x^2-1+1/(x^2+1)]dx
=x^3-x+arctanx+C
1年前
10