Answer :
We will use partial integration:[tex] \int {u} \, dv= u * v - \int {v} \, du [/tex] First we will substitute: [tex]u=e^{ \sqrt{x} } , dv = dx, du = e^{ \sqrt{x} } * \frac{1}{2 \sqrt{x} } dx, v = x[/tex]...[tex]=x * e^{ \sqrt{x} } - \int {xe^{ \sqrt{x} } \frac{1}{2 \sqrt{x} } } \, dx [/tex]
Another substitution:[tex]t= \sqrt{x} . t^{2} =x, dt= \frac{dx}{2 \sqrt{x} } [/tex]
[tex] \int {t^{2}e^{2} } \, dt =t^{2} e^{ \sqrt{x} } - \int {te^{ \sqrt{x} } } \, dt =t^{2} e ^{t} -2te^{t} +2 e^{t} [/tex]
Finally: ...=[tex]xe^{ \sqrt{x} } -xe^{ \sqrt{x} }-2 \sqrt{x} e^{ \sqrt{x} } -2e^{ \sqrt{x} } = 2( \sqrt{x} -1)*e^{ \sqrt{x} } +C[/tex]
Thank you.
Another substitution:[tex]t= \sqrt{x} . t^{2} =x, dt= \frac{dx}{2 \sqrt{x} } [/tex]
[tex] \int {t^{2}e^{2} } \, dt =t^{2} e^{ \sqrt{x} } - \int {te^{ \sqrt{x} } } \, dt =t^{2} e ^{t} -2te^{t} +2 e^{t} [/tex]
Finally: ...=[tex]xe^{ \sqrt{x} } -xe^{ \sqrt{x} }-2 \sqrt{x} e^{ \sqrt{x} } -2e^{ \sqrt{x} } = 2( \sqrt{x} -1)*e^{ \sqrt{x} } +C[/tex]
Thank you.