Question

integral x sqrt (3x^2 + 4) dx

Answer 1

try u = 3x^2

Answer 2

alright thanks

Answer 3

i mean u = 3x^2 + 4, my bad

Answer 4

alright

Answer 5

can you explain how to use u substitution for this problem

Answer 6

\[\int\limits_{}^{}x \sqrt{3x^{2} + 4} dx\]

let u = 3x^2 + 4
du = 6x dx
dx = 1/(6x) du so...

\[\frac{ 1 }{ 6 } \int\limits_{}^{} \sqrt{u} du\]

Answer 7

did I do this correctly?\[\frac{ (3x^2 + 4)^{3/2} }{9}+c\]

Answer 8

never mind

Answer 9

yeah you did it right