神马作文网求几道基础的积分题（在线等）
来源：学生作业帮 编辑：神马作文网作业帮 分类：数学作业 时间：2024/11/17 04:12:17
求几道基础的积分题（在线等）
1.∫e^(-x^2)xdx
2.∫√(1+Inx)dx/x
3.∫√(1+4sinx)cosxdx
4.∫(5x-2)dx/(x^2+4)
5.∫x^3dx/√(x^8-1)
6.∫dx/x^2+6x+13
7.∫x^5dx/(x^3-a^3)
8.∫5x-1/(x^3-3x-2)
我是自学的,所以没人问,这几题不会,麻烦大家给个大致的过程.
1.∫e^(-x^2)xdx
2.∫√(1+Inx)dx/x
3.∫√(1+4sinx)cosxdx
4.∫(5x-2)dx/(x^2+4)
5.∫x^3dx/√(x^8-1)
6.∫dx/x^2+6x+13
7.∫x^5dx/(x^3-a^3)
8.∫5x-1/(x^3-3x-2)
我是自学的,所以没人问,这几题不会,麻烦大家给个大致的过程.
1 ∫e^(-x^2)xdx
=∫1/2e^(-x^2)dx^2
=-(1/2)e^(-x^2)+c
2 ∫√(1+lnx)dx/x
= ∫√(1+lnx)dlnx
=(2/3)(1+lnx)^(3/2) +C
3 ∫√(1+4sinx)cosxdx
=∫√(1+4sinx)dsinx
=(1/4)*(2/3)(1+4sinx)^(3/2)
=(1/6)(1+4sinx)^(3/2) +C
4 原式=∫5xdx/(x^2+4)-∫2dx/(x^2+4)
=∫(5/2)/(x^2+4)dx^2-arctan(x/2)+C1
=(5/2)ln(x^2+4)-arctan(x/2)+C
5 原式=∫(1/4)/√(x^8-1)dx^4
=(1/4)ln|x^4+√[x^4+(-)a]+c
6 ∫dx/x^2+6x+13
=∫1/[(x+3)^2+4]d(x+3)
=(1/2)arctan[(x+3)/2] +C
7 原式=(1/6)*∫1/(x^3-a^3)dx^6
令t=x^3
原式=(1/6)*∫1/(t-a^3)dt^2
=(1/6)*∫2t/(t-a^3)dt
=(1/6)*∫2+2a^3/(t-a^3)dt
=(1/6)*[2t+2a^3ln(t-a^3)]+C
=∫1/2e^(-x^2)dx^2
=-(1/2)e^(-x^2)+c
2 ∫√(1+lnx)dx/x
= ∫√(1+lnx)dlnx
=(2/3)(1+lnx)^(3/2) +C
3 ∫√(1+4sinx)cosxdx
=∫√(1+4sinx)dsinx
=(1/4)*(2/3)(1+4sinx)^(3/2)
=(1/6)(1+4sinx)^(3/2) +C
4 原式=∫5xdx/(x^2+4)-∫2dx/(x^2+4)
=∫(5/2)/(x^2+4)dx^2-arctan(x/2)+C1
=(5/2)ln(x^2+4)-arctan(x/2)+C
5 原式=∫(1/4)/√(x^8-1)dx^4
=(1/4)ln|x^4+√[x^4+(-)a]+c
6 ∫dx/x^2+6x+13
=∫1/[(x+3)^2+4]d(x+3)
=(1/2)arctan[(x+3)/2] +C
7 原式=(1/6)*∫1/(x^3-a^3)dx^6
令t=x^3
原式=(1/6)*∫1/(t-a^3)dt^2
=(1/6)*∫2t/(t-a^3)dt
=(1/6)*∫2+2a^3/(t-a^3)dt
=(1/6)*[2t+2a^3ln(t-a^3)]+C