How do you use substitution to integrate #x^2*(x-2)^(1/2)#?

Question

1389 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-08T09:37:21

The answer is #=2/7(x-2)^(7/2)+8/5(x-2)^(5/2)+8/3(x-2)^(3/2)+C#

We need

#intx^ndx=x^(n+1)/(n+1)+C(n!=-1)#

Perform the substitution

#u=x-2#, #=>#, #dx=du#

Therefore,

#intx^2sqrt(x-2)dx=int(u+2)^2sqrtudu#

#=int(u^2+4u+4)sqrt(u)du#

#=int(u^(5/2)+4u^(3/2)+4u^(1/2))du#

#=intu^(5/2)du+4intu^(3/2)du+4intu^(1/2)du#

#=u^(7/2)/(7/2)+4u^(5/2)/(5/2)+4u^(3/2)/(3/2)#

#=2/7u^(7/2)+8/5u^(5/2)+8/3u^(3/2)#

#=2/7(x-2)^(7/2)+8/5(x-2)^(5/2)+8/3(x-2)^(3/2)+C#