百度作业网∫(x^4 1 x^2)
来源：学生作业帮助网 编辑：作业帮 时间：2024/06/13 20:34:09
取x=sint+1(-pi/2
x^2/[(x-3)(x+2)^2=(9/25)[1/(x-3)]+(16/25)[1/(x+2)]-(20/25)[1/(x+2)^2].原式=(9/25)∫1/(x-3)dx+(16/25)∫1/
∫(x+1)/(x^2-2x+5)dx=1/2*∫(2x-2)/(x^2-2x+5)dx+∫2/(x^2-2x+5)dx=1/2*∫[1/(x^2-2x+5)]d(x^2-2x+5)+2∫1/[(x-
(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)arctan(x
你是问为什么∫du/(u²+a²)²=(1/2a²)[u/(u²+a²)+∫du/(u²+a²)],对吗?如果是这么个问
左边=∫x√(1-x^2)dx-∫x^2dx=-1/2∫√(1-x^2)d(-x^2)-x^3/3=-1/2*2/3*(1-x^2)^(3/2)-x^3/3+C=-1/3*(1-x^2)^(3/2)-
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^2)=(1/2)ln|ln(1+x^2)|+C1x
[f(x)/f'(x)]'=[f'²(x)-f(x)f''(x)]/f'²(x)=1-f(x)f''(x)/f'²(x)因此题目中的被积函数为：[f(x)/f'(x)-f
∫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)
∫f'(x)dx/1+f^2(x)=∫df(x)/[1+f^2(x)]=arctanf(x)+c=arctan(e^x/x)+c
∫2^x*3^x/(9^x-4^x)dx=∫(2/3)^xdx/[1-(4/9)^x]=[ln(2/3)]^(-1)∫d[(2/3)^x]/{1-[(2/3)^x]^2}={[ln(2/3)]^(-1
原式=0.5∫d（x²+2x+5）/(x²+2x+5)=0.5㏑(x²+2x+5)
∵(x^4-4x^2+5x-15)/[(x^2+1)(x-2)]=[(x^4+x²-5x²-5)+(5x-10)]/[(x²+1)(x-2)]=[x²(x&su
2lnlx-4l-lnlx-3l+c∫(x-2)/(x^2-7x+12)dx=∫(x-2)/(x-4)(x-3)dx=∫[2/(x-4)-1/(x-3)]dx=∫2/(x-4)dx-∫1/(x-3)d
[(x^3-2x^2+x+1)/(x^4+5x^2+4)]=1/(x^2+1)+(x-3)/(x^2+4).原式=∫1/(x^2+1)dx+∫(x-3)/(x^2+4)dx=arctanx+(1/2)
∫(e^(x^2))x(1+x^2)dx=(1/2)∫(1+x^2)de^(x^2)=(1/2)(1+x^2).e^(x^2)-∫x.e^(x^2)dx=(1/2)(1+x^2).e^(x^2)-(1