hi ,
you might go in this way !
first , seperate the terms in this way
∫x /√(x-3) dx + 2∫ 1/ √(x-3) dx
x -3 =u
dx = du & x= u+3
2∫ 1/ √(x-3) dx = 2 u-1/2+1 / ½ = 4√(x-3) +c
∫x /√(x-3) dx = ∫ x√(x-3) / (x-3) dx
x -3 = u
x = u+3
dx = du
= ∫(u+3)√u/u du = ∫(√u + 3/√u ) du
= 2/3 (x-3)3/2 + 6 √(x-3) + k
finally,
∫(x +2)/√(x-3) dx = 2/3 (x-3)3/2 + 6 √(x-3) + 4√(x-3) + K
= 2/3 (x-3)3/2 + 10√(x-3) +K
K – arbitary constant .
Thank you , keep learning by questioning !
