TBI's question at Yahoo! Answers (Indefinite integral)

[Fernando Revilla]

Here is the question:

/ 1
| --------------------- dx
/ e^sqrt(x) + 1

Here is a link to the question:

Integral of (e^sqrt(x) + 1) with respect to x? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

Hello TBI,

Denote $I=\int(e^{\sqrt{x}}+1)\;dx$, then $I=\int e^{\sqrt{x}}\;dx+x+C$. If $t=\sqrt{x}$, then $dt=\dfrac{dx}{2\sqrt{x}}=\dfrac{dx}{2t}$ so: $$\int e^{\sqrt{x}}\;dx=2\int te^{t}\;dt$$ Using the integration by parts method:

$$\begin{aligned}\left \{ \begin{matrix}u=t\\dv=e^tdt\end{matrix}\right.& \Rightarrow \left \{ \begin{matrix}du=dt\\v=e^t\end{matrix}\right.\\& \Rightarrow \int te^{t}\;dt=te^t-\int e^t\;dt\\&=\sqrt{x}e^{\sqrt{x}}-e^{\sqrt{x}}\end{aligned}$$ As a consequence: $$\boxed{\;I=\displaystyle\int(e^{\sqrt{x}}+1)\;dx=2e^{\sqrt{x}}(\sqrt{x}-1)+x+C\;}$$