不定积分∫dx (1-2x)
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/24 05:01:34
=-1/2∫√(1-x^2)d(1-x^2)=-1/2×2/3√(1-x^2)^3+C=-1/3√(1-x^2)^3+C
∫(X^2+X+1/X)dx=x^3/3+x^2/2+lnx+C
差不多就这样再答:
∫x/(1+x^2)dx=1/2∫1/(1+X^2)d(x^2)=1/2∫1/(1+X^2)d(1+x^2)=1/2ln(1+x^2)+c
∫(arctanx)/(x^2(x^2+1))dx=∫(arctanx)/x^2dx-∫(arctanx)/(x^2+1)dx=∫(arctanx)d(1/x)-∫(arctanx)darctanx=
=∫[(x^2+1-1)/(x^2+1)]dx=∫[1-1/(x^2+1)]dx=x-arctanx
∫1/(x^2+2x+5)dx=∫1/[(x+1)^2+4]dx=∫(1/4)/[[(x+1)/2]^2+1]dx=∫(1/4)·2/[[(x+1)/2]^2+1]d((x+1)/2)=(1/2)∫1
=∫dx²/(1+x²)=ln(1+x²)+C,C为常数
很简单啊,好好观察形状就好解了
用分部积分法,(uv)'=u'v+uv',设u=ln(1+x^2),v'=1,u'=2x/(1+x^2),v=x,原式=xln(1+x^2)-2∫x^2dx/(1+x^2)=xln(1+x^2)-2∫
原式=∫(1+ln^2x)d(lnx)令lnx=u上式化为∫(1+u^2)du=u+u^3/3+c=lnx+(lnx)^3/3+c
∫x^2/√(1-x^2)dx=-∫-2x^2/2√(1-x^2)dx=-∫xd√(1-x^2)=-x√(1-x^2)+∫√(1-x^2)dx其中,解∫√(1-x^2)dx令x=sintdx=cost
∫2x/x^2+1dx=S1/(x^2+1)d(x^2+1)=ln(x^2+1)+c
∫(x^2-3x)/(x+1)dx=∫[(x+1)(x-4)/(x+1)+4/(x+1)]dx=∫(x-4)dx+∫4/(x+1)dx=x²/2-4x+4ln(x+1)+C其中C为任意常数
原式=∫dx/(2X-1)^3/2=1/2∫(2X-1)^(-3/2)d(2x-1)=-根号(2x-1)
积分:(x^2+1)/(x^4+1)dx=积分:(1+1/x^2)/(x^2+1/x^2)dx(上下同时除以x^2)=积分:d(x-1/x)/[(x-1/x)^2+(根号2)^2]=1/根号2*arc
∫(2x+1)dx=∫2xdx+∫dx=x^2+x+C