这个应该用凑微分方法会更容易的.
∫x√(3x+2) dx
=(1/3)∫3x√(3x+2) dx
=(1/3)(1/3)∫(3x+2-2)√(3x+2) d(3x)
=(1/9)∫[(3x+2)^(3/2) - 2√(3x+2)] d(3x+2)
=(1/9)*(2/5)(3x+2)^(5/2) - (1/9)*2(2/3)(3x+2)^(3/2) + C,再将答案因式分解后得
=(2/135)(9x-4)(3x+2)^(3/2) + C=(1/3)(1/3)∫(3x+2-2)√(3x+2) d(3x)=(1/9)∫[(3x+2)^(3/2) - 2√(3x+2)] d(3x+2)怎么从(3x+2-2)√(3x+2) 变到(3x+2)^(3/2) 谢谢那是次方问题√(3x+2) = (3x+2)^(1/2)，根号就是二分之一次方∴(3x+2)√(3x+2)=(3x+2)*(3x+2)^(1/2)=(3x+2)^(1+1/2)，指数定律会吧？=(3x+2)^(3/2)