作业帮∫(1,1,1)->(2,3,-4)xdx y∧2dy-z∧3dz与积分路径无关
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/22 01:11:22
我觉得三角公式会简单点,直接展开更麻烦一些1/(1-x²)=1/2*[1/(1-x)+1/(x+1)]所以展开得到1/(1-x²)^3=1/8*[-1/(x-1)+1/(x+1)]
设x=tant∫(2x^2+3)/{(x^2+1)^2}dx=∫[2tan^2(t)+3]/sec^4(t)*sec^2(t)dt=∫[2sin^2(t)+3cos^2(t)]dt=∫[2+(1+co
答：原式=1/3∫1/(3x-1)^2*d(3x-1)=1/3*(-1/(3x-1))+C=-1/(3(3x-1))+C
∫（-2,-3）1/1+xdx=ln|1+x|（-2,-3）=ln2-ln1=ln2
∫(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
设√x=t,则x=t²,dx=2tdt∴原式=3∫dt/(1+t²)=3arctant+C(C是积分常数)=3arctan(√x)+C
令x=tant,-π/2
渣渣你好.
设x=sint,dx=costdt,当x=0时,t=0,x=1,t=π/2,原式=∫[0,π/2](sint)^3(cost)^2dt=-∫[0,π/2](sint)^2(cos)^2]d(cost)
设u=(x^2-1)^(1/2),则x^2=u^2+1dx^2=d(u^2+1)=2udu∫[(x^3)/(x^2-1)^(1/2)]dx=∫[(x^2)/[2(x^2-1)^(1/2)]]dx^2=
答：∫(arctanx)^3/(1+x^2)dx=∫(arctanx)^3d(arctanx)=(1/4)(arctanx)^4+C
原式=∫dx/(cos²x+4sin²x)(应用sin²x+cos²x=1)=∫sec²xdx/(1+4sin²x/cos²x)=
令x=tany,dx=sec²ydy(1+x²)^(3/2)=(1+tan²y)^(3/2)=(sec²y)^(3/2)=sec³y原式=∫sec&s
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
（1）原式=∫[x^(2/3)+6x^(1/3)+9]dx=3/5*x^(5/3)+9/2*x^(4/3)+9x+C（2）原式=∫(4x^3-4x^2-x)dx=x^4-4/3*x^3-1/2*x^2
∫1/[(2-3x)(2x+1)]dx=∫6/[(4-6x)(6x+3)]dx=∫6/7[1/(4-6x)+1/(6x+3)]dx=1/7[∫1/(6x+3)d(6x+3)+∫1/(4-6x)d(4-