What is the Integral of e^√x/√x?

[karush]

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

ok I set $u=\sqrt{x}$ and $du=\frac{1}{2\sqrt{x}}dx$

I thot I would find a table reference for this but not sure which one could be used so now we have
$$\frac{1}{2}\int\frac{e^{u}}{u}du$$

but maybe better way

 

[MarkFL]

You have chosen your $u$-substitution well, but you haven't substituted quite correctly...think of the original integral as:

$$2\int e^{\sqrt{x}}\,\frac{1}{2\sqrt{x}}\,dx$$

Now perhaps it is more clear what your integral should look like after the substitution. :D

 

[evinda]

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

ok I set $u=\sqrt{x}$ and $du=\frac{1}{2\sqrt{x}}dx$

I thot I would find a table reference for this but not sure which one could be used so now we have
$$\frac{1}{2}\int\frac{e^{u}}{u}du$$

but maybe better way

$u=\sqrt{x}$

$du=\frac{1}{2\sqrt{x}}dx \Rightarrow dx=2 \sqrt{x} du=2udu$

So:

$$\int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx=\int \frac{e^u}{u} 2 u du=2 \int \frac{e^u}{u}u du=2 \int e^u du=2(e^u+c)=2e^{\sqrt{x}}+C$$

 

[mathbalarka]

The substitution was incorrect. $$\frac{e^{\sqrt{x}}}{\sqrt{x}} \mathrm{d}x = \frac{2}{2} \cdot \frac{e^{\sqrt{x}}}{\sqrt{x}}dx = 2e^{\sqrt{x}} d\left(\sqrt{x}\right)$$

So how would the differential form look after the u-sub $u = \sqrt{x}$?

 

[karush]

thanks finally seeing this

this best help is always here :)