- #1
cunhasb
- 12
- 0
Hoping anyone could give a hand on this...
this is what I've gotten so far...
∫(x+2)√(x-3)dx
u=(x-3)^1/2
u^2=(x-3)
x=u^2-3
dx=2u du
∫(x+2)√(x-3)dx=(u^2-3+2)*(u)*(2u) du =2∫(u^2-1)*(u)*(u)
= 2∫(u^2-1)*(u^2)
=2∫(u^4-u^2)
=(2*1/5*u^5) - ((2/5)*2*(1/3)*u^3)
=2/5*(x-3)^(5/2) - 4/15(x-3)^3/2 + K
Is it right?
Thank you so much guys...
this is what I've gotten so far...
∫(x+2)√(x-3)dx
u=(x-3)^1/2
u^2=(x-3)
x=u^2-3
dx=2u du
∫(x+2)√(x-3)dx=(u^2-3+2)*(u)*(2u) du =2∫(u^2-1)*(u)*(u)
= 2∫(u^2-1)*(u^2)
=2∫(u^4-u^2)
=(2*1/5*u^5) - ((2/5)*2*(1/3)*u^3)
=2/5*(x-3)^(5/2) - 4/15(x-3)^3/2 + K
Is it right?
Thank you so much guys...