∫|x²–3x+2|dx

∫x^3/(9+x^2)dx=1/2∫x^2/(9+x^2)dx^2(x^2=t)=1/2∫t/(9+t)dt=1/2∫(t+9-9)/(9+t)dt=1/2∫[1-9/(9+t)]dt=1/2t-9

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^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

把下边有理化,就非常容易了.再问：求详细再答：等下，我回自习室给你写下啊再答：再答：再答：再答：答案是不是它再答：应该是把3换成2是平方我看错了再问：书上的参考答案是ln[(x+3)/(x+2)]&#

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)=

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

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+

∫(x²-2x+1)/x³dx=∫(1/x-2/x²+1/x³)dx=lnx+2/x-2/x²+C

∫x^3/(1+x^2)dx=∫[x^3+x-x]/(1+x^2)dx=∫x-x/(1+x^2)dx=x²/2-1/2ln[1+x^2]+c你的好评是我前进的动力.我在沙漠中喝着可口可乐,唱

∫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

(x^2)/2-18x^(1/2)+3x+C0.5*x^2+2*x^(1/2)+C9x-2x^3+0.2*x^5+C

具体见图片内容：再问：第二步怎么来的？没认真听课现在看起来很吃力麻烦讲解下我会提高悬赏的再答：就是自然对数lnx求导的形式：(lnx)'=1/x

原式=-1/3∫e^-X^3d(-X^3)=-1/3e^-X^3+c

我想你的题应该是这样吧∫x³/(9+x²)dx=(1/2)∫x²/(9+x²)d(x²)=(1/2)∫(x²+9-9)/(9+x²

∫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

∫x^3/(9+x^2)dx=1/2∫x^2/(9+x^2)dx^2(x^2=t)=1/2∫t/(9+t)dt=1/2∫(t+9-9)/(9+t)dt=1/2∫[1-9/(9+t)]dt=1/2t-9

4/3*x^4+3/2x^2+1/4*x^2+c再问：可我算得x^4/4+(3/2)x^2+2lnlxl+c再答：你的对，我看错了

展开得到原积分=∫4^x+2*6^x+9^xdx=4^x/ln4+2*6^x/ln6+9^x/ln9+C,C为常数再问：(⊙o⊙)哦看懂了谢谢再答：不必客气的啊~

原式=∫(3x^4+3x^2-2x^2-2+2)/(x^2+1)dx=∫[3x^2-2+2/(x^2+1)]dx=x^3-2x+2arctanx+C