- #1

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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...