## Thomas' Calculus 13th Edition

$$2x{e^{\sqrt x }} - 4\sqrt x {e^{\sqrt x }} + 4{e^{\sqrt x }} + C$$
\eqalign{ & \int {\sqrt x {e^{\sqrt x }}} dx \cr & {\text{Let }}{t^2} = x,\,\,\,\,2tdt = dx,\,{\text{use the substitution}} \cr & \int {\sqrt x {e^{\sqrt x }}} dx = \int {\sqrt {{t^2}} {e^{\sqrt {{t^2}} }}} \left( {2tdt} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int {t{e^t}} \left( {2tdt} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int {2{t^2}{e^t}} dt \cr & {\text{Using the integration by parts method }} \cr & \,\,\,\,\,{\text{Let }}u = 2{t^2},\,\,\,\,du = 4tdt,\,\,\,\,\, \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = {e^t}dt,\,\,\,\,v = {e^t} \cr & \cr & {\text{Integration by parts gives}} \cr & \int {2{t^2}{e^t}} dt = 2{t^2}{e^t} - \int {{e^t}} \left( {4tdt} \right) \cr & \int {2{t^2}{e^t}} dt = 2{t^2}{e^t} - \int {4t{e^t}} dt \cr & \cr & {\text{Integrate by parts again}} \cr & \,\,\,\,\,{\text{Let }}u = 4t,\,\,\,\,du = 4dt,\,\,\,\,\, \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = {e^t}dt,\,\,\,\,v = {e^t} \cr & \int {2{t^2}{e^t}} dt = 2{t^2}{e^t} - \left( {4t{e^t} - \int {4{e^t}} dt} \right) \cr & \int {2{t^2}{e^t}} dt = 2{t^2}{e^t} - 4t{e^t} + \int {4{e^t}} dt \cr & \int {2{t^2}{e^t}} dt = 2{t^2}{e^t} - 4t{e^t} + 4{e^t} + C \cr & \cr & {\text{Write in terms of }}s{\text{, substitute }}\sqrt x {\text{ for }}t \cr & \int {\sqrt x {e^{\sqrt x }}} dx = 2{\left( {\sqrt x } \right)^2}{e^{\sqrt x }} - 4\sqrt x {e^{\sqrt x }} + 4{e^{\sqrt x }} + C \cr & \int {\sqrt x {e^{\sqrt x }}} dx = 2x{e^{\sqrt x }} - 4\sqrt x {e^{\sqrt x }} + 4{e^{\sqrt x }} + C \cr}