第1个回答 2018-04-13
第2个回答 2018-04-13
1、原式=(1/2)*∫(-∞,1) d(1+x^2)/(1+x^2)^2
=(-1/2)*1/(1+x^2)|(-∞,1)
=(-1/2)*(1/2)
=-1/4
2、原式=∫d(x^2+2x+2)/(x^2+2x+2)+∫d(x+1)/[(x+1)^2+1]
=ln|x^2+2x+2|+arctan(x+1)+C，其中C是任意常数
3、原式=∫[1/x^2-1/(1+x^2)]dx
=-1/x-arctanx+C，其中C是任意常数
4、原式=2∫(0,π/2) √[cos^3x*(1-cos^2x)]dx
=2∫(0,π/2) (cosx)^(3/2)*sinxdx
=-2∫(0,π/2) (cosx)^(3/2)d(cosx)
=(-4/5)*(cosx)^(5/2)|(0,π/2)
=(-4/5)*(-1)
=4/5本回答被提问者和网友采纳
=(-1/2)*1/(1+x^2)|(-∞,1)
=(-1/2)*(1/2)
=-1/4
2、原式=∫d(x^2+2x+2)/(x^2+2x+2)+∫d(x+1)/[(x+1)^2+1]
=ln|x^2+2x+2|+arctan(x+1)+C，其中C是任意常数
3、原式=∫[1/x^2-1/(1+x^2)]dx
=-1/x-arctanx+C，其中C是任意常数
4、原式=2∫(0,π/2) √[cos^3x*(1-cos^2x)]dx
=2∫(0,π/2) (cosx)^(3/2)*sinxdx
=-2∫(0,π/2) (cosx)^(3/2)d(cosx)
=(-4/5)*(cosx)^(5/2)|(0,π/2)
=(-4/5)*(-1)
=4/5本回答被提问者和网友采纳
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