How do you integrate #int (x^2+1)sqrt(x-2)# using substitution?

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Answer 1 · 2016-09-07T14:13:20

#= 2/105 (x-2)^(3/2) (15 x^2 + 24 x + 67)+ C#

the obvious thing is to simplify that radical so that it can be distributed

so let #z = x-2, x = z + 2, dx/dz = 1# gives us:

#implies int ((z+2)^2+1 )sqrt(z) dz#

#= int (z^2+4z + 5 )sqrt(z) dz#

#= int z^(5/2)+4z^(3/2) + 5z^(1/2) dz#

#= ( 2z^(7/2))/7+ (8z^(5/2))/5 + (10z^(3/2))/3 + C#

#=(x-2)^(3/2)( ( 2(x-2)^(2))/7+ (8(x-2))/5 + (10)/3 )+ C#

#= 2/105 (x-2)^(3/2) (15 x^2 + 24 x + 67)+ C#