x^2/(1+x^2 )dx

来源：蜘蛛抓取(WebSpider) 时间：2016-12-25 14:36 标签： xcos2xdx
求∫(x^2+1)/[(x+1)^2(x-1)]dx
原式=（1/2）∫[（x-1）/(x+1)^2+1/(x-1)]dx=(1/2)∫[(x+1-2)/(x+1)^2+1/(x-1)]dx=(1/2)∫[1/(x+1)-2/(x+1)^2+1/(x-1)]dx=(1/2)[ln(x+1)+2/(x+1)+ln(x-1)]+C=(1/2)[ln(x+1)+ln(x-1)]+1/(x+1)+C之所以拆成第一个等号,是因为设原式=∫[（ax+b）/(x+1)^2+c/(x-1)]dx,对于中括号里的部分通分,从而分子变为（a+c）x^2+(b-a+2c)x+c-b,而原来的分子是x^2+1,所以a+c=1,b-a+2c=0,c-b=1,解得a=c=1/2,b=-1/2
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扫描下载二维码不定积分∫(x^2-1)/(x^4+1)dx计算具体过程
分子分母除于x²,然后注意到分子变为 1-1/x²=(x+1/x)'然后令u=x+1/x,可以得到答案参考资料中有详细过程
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好多都忘了，应该是（x^2-1）/(x^4+1)
=x^2/(x^4+1)-1/(x^4+1)
=1/(x^2+1/x^2)-1/(x^4+1)
=(x^2+1/x^2)的-1次方-（x^4+1）的-1次方 现在减号两边积分，你看行不行。
对被积函数分子分母同时除以x^2,得原式=∫[1-(1/x^2)]/[x^2+(1/x^2)]dx=∫1/{[x+(1/x)]^2-2}d[x+(1/x)]【凑微分法】=[(√2)/4]ln|[x+(1/x)-√2]/[x-(1/x)+√2]|+C【常用公式】=[(√2)/4]ln|[x^2-√2x+1]/[x^2+√2x+1]|+C
扫描下载二维码高数,解/[(x^2+1)(x^2+x+1)]怎样得出或想出,∫[-x/(x^2+1)+(x+1)/(x^2+x+1)]dx,用待定系数法求不出额.
待定系数法可以啊.1/[(x²+1)/(x²+x+1)] = (Ax+B)/(x²+1) + (Cx+D)/(x²+x+1)1 = (Ax+B)(x²+x+1) + (Cx+D)(x²+1)1 = Ax³+Ax²+Ax + Bx²+Bx+B + Cx³+Cx + Dx²+D1 = (A+C)x³ + (A+B+D)x² + (A+B+C)x + (B+D)A+C = 0 ...(i)A+B+D = 0 ...(ii)A+B+C = 0 ...(iii)B+D = 1 ...(iv)将(i)代入(iii)得出B = 0 ...(v)将(iv)代入(ii)得出A = -1 ...(vi)将(v)(vi)代入(iii)、(ii)得出C = D = 11/[(x²+1)/(x²+x+1)] = -x/(x²+1) + (x+1)/(x²+x+1)这个积分的答案,有需要就看看吧,不过你自己应该也会做的.-∫x/(x²+1) dx + ∫(x+1)/(x²+x+1) dx令x+1 = E(2x+1)+Fx+1 = 2Ex + E + F2E = 1 => E = 1/2E + F = 1 => F = 1/2= -∫d(x²/2)/(x²+1) + ∫[(1/2)(2x+1)+1/2]/(x²+x+1) dx= -(1/2)∫d(x²+1)/(x²+1) + (1/2)∫(2x+1)/(x²+x+1) dx + (1/2)∫dx/(x²+x+1)= -(1/2)ln|x²+1| + (1/2)∫d(x²+x+1)/(x²+x+1) + (1/2)∫d(x+1/2)/[(x+1/2)²+3/4]= -(1/2)ln|x²+1| + (1/2)ln|x²+x+1| + (1/2)*√(4/3)*arctan[(x+1/2)*√(4/3)] + C= (1/2)ln|(x²+x+1)/(x²+1)| + (1/√3)arctan[(2x+1)/√3] + C = (1/2)ln|1+x/(x²+1)| + (1/√3)arctan[(2x+1)/√3] + C
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扫描下载二维码求积分：∫[x^2/(x-1)^100]dx.得的结果和答案总是不一样,就是用的分步积分.
∫[(x^2- 1+1)/(x-1)^100]dx=∫[(x+1)/(x-1)^99]dx+∫[1/(x-1)^100]dx=∫[(x - 1+2)/(x-1)^99]dx+∫[1/(x-1)^100]dx=∫[1/(x-1)^98]dx+∫[2/(x-1)^99]dx+ ∫[1/(x-1)^100]dx=(- 1/97)[1/(x-1)^97]+(- 2/98)[1/(x-1)^98]+(- 1/99)[1/(x-1)^99]+C
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∫x^2dx/(x-1)^100=∫(x+1)dx/(x-1)^99+∫dx/(x-1)^100=∫dx/(x-1)^98+∫2dx/(x-1)^99+∫dx/(x-1)^100=(-1/97)*(1/(x-1)^97)+(-1/49)*(1/(x-1)^98)-(1/99)(1/(x-1)^99)+C ∫x^2dx/(x-1)^100=(-1/9...
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