Question

数学，1 2 3 4题，求过程，谢谢。。。

Answer 1

1.  解：令x^(1/6)=t,那么x=t^6，dx=6t^5dt，代入得：
   ∫(x^(1/3)/x(√x+x^(1/3))dx = ∫(t^(2)/t^6(t^3+t^2))6t^5dt=∫(1/(1+t^2))dt=arctant+C
                    =arctanx^(1/6)+C

2. ∫1/2+√2x+1dx令√2x+1=t  dx=dt

3. 令x=cost
   则原式=∫√（1+cost）/(1-cost)dcost
   =∫√（1-cos^2t）/(1-cost)^2dcost
   ∫-sin^2t/(1-cost)dt
   =∫（cos^2t-1）/(1-cost)dt
   =∫-cost-1dt
   =sint-t+c
   还原

4. 原式＝∫[2dx/√[1-(1-2x)²]=-∫[d(1-2x)]/√[1-(1-2x)²]=-arcsin(1-2x)+C.