∫dxcos^3x

∫{cos(x/2)[cos(x/2+sin(x/2)]}dxcos*sin有公式么?2sinx*cosx=sin2x(cosx)^2-(sinx)^2=cos2x.由第二个又可以推出1-2(sinx)^2=cos2x2(cosx)^2-1

∫dxcos(x)dx用matlab该怎样表达,打比方有个式子∫cos(x)dx下限0上限2我知道求这种式子用matlab应该这么表达symsxint(cos(x),x,0,2)现在在式子前面乘以一个dx∫dxcos(x)dx下限0上限2请

∫[(x^2-x+6)/(x^3+3x)]dx(x^2-x+6)/(x^3+3x)=2/x-(x+1)/(x^2+3).原式=∫2/xdx-∫(x+1)/(x^2+3)dx=2ln|x|-(1/2)ln(x^2+3)-(1/√3)arcta

∫(x^3-x^2+x+1)/(x^2+1)dx∫(x+4)/(x^2-x-2)dx1.先化为x-1+2/(x^2+1)再积分,=x^2乘以1/2-x+2乘以arctanx+c2.先化为(x+4)/（x-2)（x+1)=A/(x-2)+B/

求解∫(x^2+3x+2)/(x^3+2x^2+2x)dx

计算∫（x^4-2x^3+x^2+1）/x（x-1)²dx答案见附图∫(x^4-2x^3+x^2+1)dx/[x(x-1)^2]=∫(x^4-2x^3+x^2)dx/(x^3-2x^2+x)+∫[x-(x-1)]dx/[x(x-1

∫[f(x)/f’(x)-f^2(x)f’’(x)/f’^3(x)]dx第一步通分.之后[f(x)/f'(x)]df(x)/f'(x)=1/2[fx/f'x]^2我写的可能简单勒点.不过应该能看懂吧.嘿嘿.加油∫[f(x)/f’(x)-f^

∫[f(x)/f'(x)-f^2(x)f"(x)/f'^3(x)]dx如题[f(x)/f'(x)]'=[f'²(x)-f(x)f''(x)]/f'²(x)=1-f(x)f''(x)/f'²(x)因此题目中的被积

设f(x)=x㏑(1+x^2),x≥0.(x^2+2x-3)e^(-x),x＜0,求∫f(x)dx当x=0时,f(x)不连续,故f(x)的原函数分成两部分：x>0,∫f(x)dx=∫x㏑(1+x^2)dx=(1/2)∫㏑(1+x^2)d(x

∫(3x+2)/(x(x+1)^3)dx原式=∫[2/x-2/(x+1)-2/(x+1)²+1/(x+1)³]dx=2ln│x│-2ln│x+1│+2/(x+1)-(1/2)/(x+1)²+C(C是积分常数)=

∫x/（x^2+3x+3）dx∫x/（x^2+3x+3）dx=∫(x+3/2-3/2)/（x^2+3x+3）dx=∫(x+3/2)/（x^2+3x+3）dx-3/2∫dx/（x^2+3x+3)=1/2∫d(x^2+3x+3)/（x^2+3x

∫2x²+3x-5/x+3dx设x+3＝t→dx＝dt,代入原式得∫[(2x²+3x-5)/(x+3)]dx＝∫[(2(t-3)²+3(t-3)-5)/t]dt＝∫[2t+(4/t)-9]dt＝t²+

∫(x^3+3x+2/x)dx4/3*x^4+3/2x^2+1/4*x^2+c

∫(x^3+1)/(x(1-x^3))dx(1+x³)/[x(1-x³)]=(1+x³)/[x(1-x)(1+x+x²)]令(1+x³)/[x(1-x)(1+x+x²)]=A/(1

∫(2x-3)/(x^2-3x+8)dxln(x^2-3x+8)-2/(根号23)arctan[(2x-3)/根号23]过程写起来就太麻烦了,我说下思路把先把2x-3拆成2x跟-3,变成两个积分2x那部分就是ln(x^2-3x+8)后面一半

∫(3x+2)/x(x+1)^3dx求原函数是吗?若是则∫(3x+2)/x(x+1)^3dx=∫[2/x-2/(x+1)-2/(x+1)²+1/(x+1)³]dx=2ln│x│-2ln│x+1│+2/(x+1)-(1/2

∫（x^3-x)(3x^2-1)dx（x^3-x)(3x^2-1)=3x^5-4x^3+x所以积分就是：（3/6)x^6-(4/4)x^4+(1/2)x^2化简就是：（1/2)x^6-x^4+(1/2)x^2

不定积分∫(x^5+x^4-8)/x^3-xdx∫(x^5+x^4-8)dx/(x^3-x)=∫(x^5-x^3+x^4-x^2+x^3-x+x^2+x-8)dx/(x^3-x)=∫(x^2+x+1)dx+∫(x^2+x-8)dx/(x^3

∫x+3分之x²+7x+12dx∫(x²+7x+12)/(x+3)dx=∫(x+3)(x+4)/(x+3)dx=∫(x+4)dx=x平方/2+4∫(x+x²/3+7x+12)dx=∫xdx+(1/3)∫x^2d

∫(3x-2)/(x^2-2x+5)dx