∫3x^2╱1 x^3dx

∫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

∫（3x^2-2x+2)dx=x^3-x^2+2x+C∫(2x-1)^2dx=∫4x^2-4x+1dx=4*x^3/3-4*x^2/2+x+C=4/3*x^3-2x^2+x+C

(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

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|+

∫(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

上下乘以X^2再积分再问：具体点再答：x^2/（x^3（1+x^3））dx=1/3*（1/（x^3（1+x^3）））dx^3=1/3（1/（t（1+t）））dt=1/3（1/t-1/（1+t））dt=

∫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

令x=tant,-π/2

第一步就错了,后面错得更离谱.分子从x变成1-u而不是1.于是int(x/(1-x)^3)dx=int((1-u)/u^3)d(-u)=int(u-1)/u^3du=int1/u^2du-1/u^3d

答：∫(arctanx)^3/(1+x^2)dx=∫(arctanx)^3d(arctanx)=(1/4)(arctanx)^4+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^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^2-3x)/(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

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