Verified answer

Put u = √(x). Then x = u², dx = 2u du, and e^(√x) = e^u. Upon substitution, the integral becomes

∫ 2u^3 e^u du.

Now, integrate by parts (it'll take a few steps.)

Like what Lake R pointed, make the substitution z = √x. I use z instead of u to aviod confusion when integrating by parts later.

z = √x <==> z² = x

dz = dx/(2√x)

dz = dx/(2z)

2z dz = dx

∫ x*e^(√x) dx

= ∫ z²*e^(z) * (2z dz)

= 2 ∫ z³e^(z) dz

Let u = z³ and dv = e^(z) dz, then du = 3z² dz and v = e^(z). Integrating by parts (∫ u dv = u v - ∫ v du) yields

= 2[z³*e^(z) - ∫ 3z²*e^(z) dz]

= 2z³*e^(z) - 6 ∫ z²*e^(z) dz

Now, let u = z² and dv = e^(z) dz, then du = 2z and v = e^(z). Integrating by parts for the second time gives

= 2z³*e^(z) - 6[z²*e^(z) - ∫ 2z*e^(z) dz]

= 2z³*e^(z) - 6z²*e^(z) + 12 ∫ z*e^(z) dz

This time, let u = z and dv = e^(z) dz, then du = dz and v = e^(z). Integrating by parts for the third time yields

= 2z³*e^(z) - 6z²*e^(z) + 12[z*e^(z) - ∫ e^(z) dz]

= 2z³*e^(z) - 6z²*e^(z) + 12z*e^(z) - 12 ∫ e^(z) dz

= 2z³*e^(z) - 6z²*e^(z) + 12z*e^(z) - 12e^(z) + c

= 2e^(z) [z³ - 3z² + 6z - 6] + c

= 2e^(z) [x³/² - 3x + 6√x - 6] + c

If this helps, choose Lake R's answer as the best.