计算下列不定积分,那个S拉长的符号不知道怎么打(x+2)/(x^2-2x+4)^1/2 dx1/(1+x^2)^2 dx1/(3+(sinx)^2) dx1/(sin(x^1/2))^2 dx(x-cosx)/(1+sinx) dx

Question

计算下列不定积分,
那个S拉长的符号不知道怎么打
(x+2)/(x^2-2x+4)^1/2 dx
1/(1+x^2)^2 dx
1/(3+(sinx)^2) dx
1/(sin(x^1/2))^2 dx
(x-cosx)/(1+sinx) dx

Answer 1

1、
∫ (x+2)dx/√(x²-2x+4)
= (1/2)∫ (2x-2+6)dx/√(x²-2x+4)
= (1/2)∫ (2x-2)dx/√(x²-2x+4) + 3∫ dx/√(x²-2x+4)
= (1/2)∫ d(x²-2x+4)/√(x²-2x+4) + 3∫ dx/√[(x-1)²+3]
= (1/2)*2√(x²-2x+4) + 3∫ d(x-1)/√[(x-1)²+(√3)²]
= √(x²-2x+4) + 3ln|x-1+√[(x-1)²+3]| + C
2、
∫ dx/(1+x²)²,u=tanθ,du=sec²θdθ
= ∫ sec²θdθ/sec^4θ
= ∫ cos²θ dθ
= (1/2)∫ (1+cos2θ) dθ
= (1/2)(θ+1/2*sin2θ) + C
= (1/2)[x/(1+x²)+arctanx] + C
3、
∫ dx/(3+sin²x)
= ∫ dx/[3+(1-cos2x)/2]
= 2∫ dx/(7-cos2x),y=2x,dy=2dx
= ∫ dy/(7-cosy),u=tan(y/2),dy=2du/(1+u²),cosy=(1-u²)/(1+u²)
= ∫ du/(3+4u²)
= (1/4)∫ du/(u²+3/4)
= (1/4)∫ du[u²+√(3/4)²]
= (1/4)*[1/√(3/4)]*arctan[u/√(3/4)] + C
= (1/4)*(2/√3)*arctan(2u/√3) + C
= (1 / 2√3)*arctan(2tan(x/2)/√3) + C
= [arctan(2tanx / √3)] / (2√3) + C
4、
∫ dx/sin²√x,u=√x,dx=2u du
= 2∫ ucsc²u du
= -2∫ u d(cotu)
= -2ucotu + 2∫ cotu du
= -2ucotu + 2∫ 1/sinu dsinu
= -2ucotu + 2ln|sinu| + C
= 2ln|sin√x| - 2√x*cot√x + C
5、
∫ (x-cosx)dx/(1+sinx)
= ∫ (x-cosx)(1-sinx)dx/[(1+sinx)(1-sinx)]
= ∫ (x-cosx-xsinx+sinxcosx)dx/cos²x
= ∫ xsec²x dx -∫ secx dx - ∫ xsecxtanx dx + ∫ tanx dx
= ∫ x dtanx - ∫ secx dx - ∫ x dsecx + ∫ tanx dx
= (xtanx - ∫ tanx dx) - ∫secx dx - (xsecx - ∫ secx dx) + ∫ tanx dx
= xtanx - xsecx + C
= x(tanx - secx) + C