作业帮积分的题~somebody help me~
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/11/06 04:24:19
积分的题~somebody help me~
∫cos(x)/sin^2(x)dx这个怎么积分出来~
还有∫（3x+2)^2+(1/(3x+2)^2)dx
∫3sin(2x+1)+4/(2x+1)
∫x(x^2+1)^三分之二
∫1/（x(x+1))
∫cos(x)/sin^2(x)dx这个怎么积分出来~
还有∫（3x+2)^2+(1/(3x+2)^2)dx
∫3sin(2x+1)+4/(2x+1)
∫x(x^2+1)^三分之二
∫1/（x(x+1))
∫cos(x)/sin^2(x)dx
d/dxsinx=cosx
所以∫cos(x)/sin^2(x)dx=-1/3sin^3(x)+C
∫（3x+2)^2+(1/(3x+2)^2)dx
=6x+4-1/9(3x+2)^3+C
∫3sin(2x+1)+4/(2x+1)dx
=[3cos(2x+1)]/2+2ln(2x+1)+C
∫x(x^2+1)^(2/3)dx
因为d/dx(x^2)=2x
所以∫x(x^2+1)^(2/3)dx=[3(x^2+1)^(5/3)]/10+C
∫1/（x(x+1))dx
1/(x(x+1))=1/x-1/(x+1)
所以∫1/（x(x+1))dx=lnx+ln(x+1)+C
d/dxsinx=cosx
所以∫cos(x)/sin^2(x)dx=-1/3sin^3(x)+C
∫（3x+2)^2+(1/(3x+2)^2)dx
=6x+4-1/9(3x+2)^3+C
∫3sin(2x+1)+4/(2x+1)dx
=[3cos(2x+1)]/2+2ln(2x+1)+C
∫x(x^2+1)^(2/3)dx
因为d/dx(x^2)=2x
所以∫x(x^2+1)^(2/3)dx=[3(x^2+1)^(5/3)]/10+C
∫1/（x(x+1))dx
1/(x(x+1))=1/x-1/(x+1)
所以∫1/（x(x+1))dx=lnx+ln(x+1)+C
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