Integrate $\int e^{\sqrt{x}}dx$

Substitute $x=y^2$. Then $dx=2ydy$. Hence
\[(1)               \int e^{\sqrt{x}}dx=\int e^{\sqrt{y^2}}2ydy=2\int ye^{y}dy.\]
Now we try to integrate $\int ye^{y}dy$ using integration by parts. Let 
\[u=y \ \ and \ \ dv=e^{y}     \Rightarrow   du=dy \ \ and    v=e^{y}.\]
Using the integrarion by parts formula:
\[\int u dv=uv-\int v du,\]
we have
\[(2)           \int ye^{y}dy=\int u dv=uv-\int v du=y e^{y}-\int e^{y}dy=y e^{y}-e^{y}.\]
We could add a constant in the above, but we can just add it at the last step. So putting (1) and (2) together and recalling that $y=\sqrt{x}$ we get
\[ \int e^{\sqrt{x}}dx=2(ye^{y}-e^{y})+c=2(\sqrt{x}e^{\sqrt{x}}-e^{\sqrt{x}})+c.\]