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))
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![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))](/uploads/image/z/9655608-48-8.jpg?t=%E7%A7%AF%E5%88%86%E7%9A%84%E9%A2%98%7Esomebody+help+me%7E%E2%88%ABcos%28x%29%2Fsin%5E2%28x%29dx%E8%BF%99%E4%B8%AA%E6%80%8E%E4%B9%88%E7%A7%AF%E5%88%86%E5%87%BA%E6%9D%A5%7E%E8%BF%98%E6%9C%89%E2%88%AB%EF%BC%883x%2B2%29%5E2%2B%281%2F%283x%2B2%29%5E2%29dx%E2%88%AB3sin%282x%2B1%29%2B4%2F%282x%2B1%29%E2%88%ABx%28x%5E2%2B1%29%5E%E4%B8%89%E5%88%86%E4%B9%8B%E4%BA%8C%E2%88%AB1%2F%EF%BC%88x%28x%2B1%29%29)
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))
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))
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
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
晕,别听一楼的.....
∫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(2...
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晕,别听一楼的.....
∫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
收起
我一点都看不懂,汗汗汗
1、换元,把x换成sinx
2、同上,把x换成3x+2
3、同上,把x换成2x+1
4、同上,把x换成x^2+1
5、变成∫[1/x-1/(x+1)]dx