Question

大一高数 括号3和4 过程详细点 谢谢

Answer 1

（3）∫dx/[(x+1)^2*(x-1)^4]^(1/3) 令x-1=t^3，则dx=3t^2dt
=∫3t^2dt/[(t^3+2)^2*(t^3)^4]^(1/3)
=∫3t^2dt/[(t^3+2)^(2/3)*t^4]
=∫3dt/[(t^3+2)^(2/3)*t^2]
=∫3t^2dt/[(t^3+2)^(2/3)*t^4]
=∫3d[(t^3+2)^(1/3)]/t^4
=3(t^3+2)^(1/3)]/t^4-3∫[(t^3+2)^(1/3)]*(-4)*1/t^5dt
=3(t^3+2)^(1/3)]/t^4+12∫[(t^3+2)^(1/3)]*1/t^5dt
=3(t^3+2)^(1/3)]/t^4+12∫[(1+2/t^3)^(1/3)]/t^4dt
=3(t^3+2)^(1/3)]/t^4-2∫[(1+2/t^3)^(1/3)]d(1+2/t^3)
=3(t^3+2)^(1/3)]/t^4-2*3/4*[(1+2/t^3)^(4/3)]+C
=3(x+1)^(1/3)/(x-1)^(4/3)-3/2*[(x+1)/(x-1)]^(4/3)+C
（4）∫xdx/[(3x+1)^(1/2)+(2x+1)^(1/2)] 分母有理化：
=∫x[(3x+1)^(1/2)-(2x+1)^(1/2)]dx/[(3x+1)-(2x+1)]
=∫[(3x+1)^(1/2)-(2x+1)^(1/2)]dx
=2/3*(3x+1)^(3/2)*1/3-2/3*(2x+1)^(3/2)*1/2+C
=2/9*(3x+1)^(3/2)-1/3*(2x+1)^(3/2)+C