How do you evaluate the integral #int xsqrt(x-2)#?

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Answer 1 · 2017-01-29T08:46:03

The answer is #=2/15(x-2)^(3/2)(3x+4)+ C#

We need

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

We solve this integral by substitution

Let #u=x-2#

#du=dx#

and #x=u+2#

Therefore,

#intxsqrt(x-2)dx=int(u+2)sqrtu du#

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

#=u^(5/2)/(5/2)+2*u^(3/2)/(3/2)#

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

#=2/15u^(3/2)(3u+10)#

#=2/15(x-2)^(3/2)(3x-6+10)+C#

#=2/15(x-2)^(3/2)(3x+4)+ C#