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Question Number 65665 by mathmax by abdo last updated on 01/Aug/19

$$calculate{\int }_{-2}^{+\mathrm{\infty }}\frac{e^{-x}}{\sqrt{x+2}}dx$$$$$$

Commented by mathmax by abdo last updated on 02/Aug/19

$$letI={\int }_{-2}^{+\mathrm{\infty }}\frac{e^{-x}}{\sqrt{x+2}}dxchangement\sqrt{x+2}=tgivex+2=t^2\Rightarrow$$$$I={\int }_0^{+\mathrm{\infty }}\frac{e^{-(t^2-2)}}{t}\left(2t\right)dt=2{\int }_0^{+\mathrm{\infty }}{e}^{-t^2+2}dt=e^2{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-t^2}dt$$$$=e^2\sqrt{\pi }$$

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