Please jawab. Pake jalan ya bingung :((

Int (3x √(3x^2 + 1)) dx
Misal
u = 3x^2 + 1
du = 6x dx => dx = du/(6x)
Int (3x √u) du/(6x)
= Int 1/2 u^(1/2) du
= (1/2)/(3/2) u^(3/2) | batas 0 sampai 1
= 1/3 u √u | batas 0 sampai 1
= 1/3 (3x^2 + 1) √(3x^2 + 1) | batas 0 sampai 1
= 1/3 (3(1)^2 + 1) √(3(1)^2 + 1) - 1/3 (3(0)^2 + 1) √(3(0)^2 + 1)
= 1/3 (4) √4 - 1/3 (1) √1
= 8/3 - 1/3
= 7/3