积分(2x^2 1) (4x^2 1)
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![积分(2x^2 1) (4x^2 1)](/uploads/image/f/6585063-15-3.jpg?t=%E7%A7%AF%E5%88%86%282x%5E2+1%29+%284x%5E2+1%29)
∫(x^2+2x-3)dx/x^4=∫dx/x^2+2∫dx/x^3-3∫dx/x^4=-1/x+2*1/(-3+1)*x^(-3+1)-3*1/(-4+1)*x^(-4+1)+C=-1/x-1/x^
见图,我觉得应该是对的,你自己再看看过程哈,我敢保证方法是对的
1/(x^2-4x)=(-1/4)x+(1/4)(1/(x-4))∫1/(x^2-4x)dx=(-1/4)∫[1/x-1/(x-4)]dx=(-1/4)[ln|x|-ln|x-4|]+C
答:∫[(x^4)/(1+x^2)]dx=∫{[(x^2-1)(x^2+1)+1]/(1+x^2)}dx=∫(x^2-1)+1/(1+x^2)dx=(x^3)/3-x+arctanx+C
∫dx*(secx)^2=∫dx/(cosx)^2=∫dx[(sinx)^2+(cosx)^2]/(cosx)^2=∫(sinx)^2/(cosx)^2dx+∫dx=∫sinx(-d(cosx))/(
被积函数分子分母除以x²有∫(x^2+1)/(x^4+1)dx=∫(1+1/x²)/(x²+1/x²)dx令u=x-1/x,则du=(1+1/x²)d
可惜,楼上解繁了,也解错了.利用待定系数法:∫dx/(x²+1)x⁴=∫[A/x+B/x²+C/x³+D/x⁴+(Ex+F)/(x²+1
过程很简单,用第二类换元积分法便可解决请看图:
写出原函数x^3/3-2.5x^2+4x把4和1代入做差
∫x/[(x^2+1)(x^2+4)]dx=1/3∫x[1/(x^2+1)-1/(x^2+4)]dx=1/3[∫x/(x^2+1)dx-∫x/(x^2+4)dx]=1/3[1/2∫1/[(x^2+1)
(4x^2/(3x^2+2))dx=(4/3)+(8/9)/(x^2+(2/3))dx积分得(4/3)x+(8/9)(√3/√2)arctan[(√3/√2)x]+C(x^5/(根号x^3+1))=(
∫x^2/(x^2+4)dx=∫1-4/(x^2+4)dx=∫1-1/[(x/2)^2+1]dx=x-∫2/[(x/2)^2+1]d(x/2)=x-2arctan(x/2)+C,C为常数
原式=4arctanx|(0,1)-1/2ln(1+x^2)|(0,1)=4×π/4-1/2×(ln2-ln1)=π-1/2ln2
解∫1/(1-x)²dx=-∫1/(1-x)²d(1-x)=-∫1/u²du=-(-1/u)+C=1/u+C=1/(1-x)+C
4/1怎么回事啊
∫dx/(根号5-4x-x^2)=积分1/根号(3^2-(x+2)^2)d(x+2)=1/3积分1/根号(1-[(x+2)/3]^2)d(x+2)=积分1/根号(1-[(x+2)/3]^2)d[(x+
答案有错,∫x^2/(1-x^4)dx=∫x^2/(1-x^4)dx=∫x^2/[(1+x^2)(1-x^2)]dx=∫1/2[1/(1-x^2)-1/(1+x^2)]dx=1/2∫[1/(1-x^2
尝试下把X换做tanB,不保证能做出来
有分部积分知识可知:∫x(lnx)²dx =(1/2)∫(lnx)²d(x²)=x²(lnx)²/2—∫xlnxdx=x²(lnx)
设x=tantx的4次方为tant的四次1+tant的方为sect的平方后面应该都会吧?sect=1/cost就是sintcost的平方啊,不懂可以继续追问,不过我想你应该会的