1除以2x平方减1的积分-搜答案-智慧作业帮

xdx/(1-x*x)^(1/2)=-1/2*d(1-x*x)/(1-x*x)^(1/2)再问：我也是这样算的最后是负一但答案是1

∫xdx/√(1-x²)=(1/2)∫2xdx/√(1-x²)=(1/2)∫dx²/√(1-x²)=-(1/2)∫d(-x²)/√(1-x²

1、由于被积函数是奇函数,积分区间是对称区间,因此结果为0.2、∫[0--->a](a²-x²)^(5/2)dx换元法：令x=asint,(a²-x²)^(5/

设x=sint,dx=costdt,(以下省略积分符号）原式=[(sint)^2/cost]costdt=(sint)^2dt=(1-cos2t)/2*dt=1/2[dt-cos2tdt)=1/2t-

5/(x²+x)-1/(x²-x)=05/x(x+1)-1/x(x-1)=05x-5-x-1=04x=6x=3/2x=2分之3再问：5x-5-x-1=0这个是怎么算的？再答：方程两

((x+2)/(x^2-2x)-(x-1)/(x^2-4x+4))/((x^2-16)/(x^2+4x))=((x+2)(x-2)-(x-1)x)/((x-2)^2*x)/((x-4)/x)=((x-

不易输入,发个图片：

二分之根号2乘以arctan[(x-1)/根号(2x)]+四分之根号2乘以lnabs[(x+根号2x+1)/(x-2x+1)]+C

再问：非常感谢您的指点。

1除以(x的平方+3x+2)+1除以(x平方+5x+6)+1除以(x平方+7x+12)=1/（x+1）（x+2）+1/(x+2)(x+3)+1/(x+3)(x+4)=1/(x+1)-1/(x+2)+1

你的题是[(x-2)/(x+2)]-1=3/(x^2-4)等式前面通分得：（x-2)/(x+2)-(x+2)/(x+2)=3/(x^2-4)化简：-4/(x+2)=3/(x^2-4)3x+6=-4x^

先作换元,令：t=arctanx,则x=tant所求积分=s[0,Pi/4](Ln[1+tant])dt=s[0,Pi/4](Ln[1+tan(Pi/4-t)])dt(#)=s[0,Pi/4](Ln[

62/3具体看图.可能有错.

亲,记得采纳哦.

先说答案x1=2x2=1/22(x*2+1/x*2)-3(x+1/x)-1=02(x+1/x)*2-4-3(x+1/x)-1=0(x+1/x)*2-3/2(x+1/x)-5/2=0(x+1/x-3/4

Unexpectedlyonlymecanhelpyou?Don'tmindIsayEnglish.LetN=∫(e→+∞)f(x)dx,sincethisintegralisconvergent,i

(x+1)/x^2-2x^2/(x+1)=1令a=(x+1)/x^2,则x^2/(x+1)=1/aa-2/a-1=0a^2-a-2=0(a-2)(a+1)=0a=2,a=-1a=2(x+1)/x^2=

2/(x^2-1)-1/(x-1)=12/(x+1)(x-1)-1/(x-1)=1两边乘(x+1)(x-1)2-（x+1）=x^2-1x^2+x-2=0(x+2)(x-1)=0x1=-2x2=1(舍)

∫(1→3)(x³+1/x²)dx=x⁴/4-1/x：[1→3]=(3⁴/4-1/3)-(1/4-1)=62/3