作业帮∫(2x+2)/[(x-1)*(x^2+1)^2]dx
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/30 03:12:37
![∫(2x+2)/[(x-1)*(x^2+1)^2]dx](/uploads/image/z/1983651-51-1.jpg?t=%E2%88%AB%282x%2B2%29%2F%5B%28x-1%29%2A%28x%5E2%2B1%29%5E2%5Ddx)
x_KPƿ msn
}P!ŹbBR/ ôL/ ,.9ۮ
mJ- f{yΏÑ~BCNc
E!(đV8OKl8Xwv "_r6 ϰ9C{DwpqoNFVzttgdFaު7jv,ƞɸAE[&
}7e0S}S D
∫(2x+2)/[(x-1)*(x^2+1)^2]dx
∫(2x+2)/[(x-1)*(x^2+1)^2]dx
∫(2x+2)/[(x-1)*(x^2+1)^2]dx
令(2x+2)/[(x-1)(x²+1)²]=A/(x-1) + (Bx+C)/(x²+1) + (Dx+E)/(x²+1)²
右边通分合并后,与左边比较系数得：
A=1,B=-1,C=-1,D=-2,E=0,
因此：(2x+2)/[(x-1)(x²+1)²]=1/(x-1) - (x+1)/(x²+1) - 2x/(x²+1)²
原积分=∫ 1/(x-1) dx - ∫ x/(x²+1) dx - ∫ 1/(x²+1) dx - 2∫ x/(x²+1)² dx
=ln|x-1| - (1/2)ln(x²+1) - arctanx - ∫ 1/(x²+1)² d(x²)
=ln|x-1| - (1/2)ln(x²+1) - arctanx + 1/(x²+1) + C
若有不懂请追问,如果解决问题请点下面的“选为满意答案”.
*-----------------------------------------------*| 6 4 X | 8 X X | X X 5 || X X X | X X X | X 7 8 || X X X | X X X | X X X ||---------------+---------------+--------------- || X X X | X X X | 5 1 X || X X X | X 6 X | X X X || 8 X X | 3 5 X | 2 X X ||
填九宫格帮帮忙.x x 6 x x 7 x x 98 x x x 3 x 1 x x 9 x x 6 x 5 x 3 x x x 3 x x x x 1 8x x x 9 x 1 x x x2 1 x x x x 6 x x x 6 x 7 x 3 x x 1 x x 9 x 2 x x x 47 x x 8 x x 5 x x
七年级下册政治复习提纲(山东人民出版社)第五单元是 青春的脚步 青春的气息别弄错了啊!格式：X X X X X X X X X 1、X X X X X X X X X X.2、X X X X X X X X X.3、X X X X X X X X X X X.
分解因式----（X*X-X)(X*X-X-2)+1
】x(x+1)+x^2(x-1)
1/(1-x)+1/(1+x)+2/(1+x*x)+4/(1+x*x*x*x)+8/(1-x*x*x*x*x*x*x*x)怎么做
x-1/x-2
{2X+1}+{X}
(x^2-1)/x
x^2-1/x
2X+|X+1|
｜x-1｜+｜x-2｜
(x+1)/(x+2)
/X+1/+/X+2/
2x/(1-x)
|X+2|+|X-1|
2x-1 (x
|x-2|+|x-1|