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

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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: $$\boxed{\;I=\displaystyle\int(e^{\sqrt{x}}+1)\;dx=2e^{\sqrt{x}}(\sqrt{x}-1)+x+C\;}$$