Integral Calculus......... ò e^x(3e^x + 1)^1/2 dx =?

Question

= (1/5)(3ex + 1)5/3 + C ??? is this correct?

Answer 1

u=e^x
du=e^x dx
then
integral of e^x(3e^x + 1)^1/2 dx = integral of (3u+1)^1/2 du
= 1/3 (3u+1)^3/2 (2/3)+C
=2/9 (3e^x+1)^3/2 +C
so your answer is wrong

Answer 2

u = 3e^x, du = 3e^x dx => integral (1/3)(u+1)^(1/2)
= (1/3)(u+1)^(3/2) / (3/2) + C
= 2/9(u+1)^(3/2) + C
= 2/9(3e^x + 1)^(3/2) + C