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

Question

1591 views around the world

You can reuse this answer
Creative Commons License

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:

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

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

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

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

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

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

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