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The value of the integral $$ \int_{-2}^0 \frac{x}{\sqrt{e^x+\left(x+2\right)^2}} \,dx,$$ is: (a) $-1 \,\,$ (b)$-2 \,\,$ (c) $-e\,\,$ (d) $2 - e\,\,$ (e) another answer

(yes, this is one of the options)

I am having trouble solving this question. It seems like this integral has no elementary primitive and wolfram can offer only a numeric value which does not satisfy the given answers. Maybe this exercise is wrong or something but I am not sure, it is after all for an examination, so I do not know if they made a mistake or not.

Zacky
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Jon9
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    numerical evidence suggests it is equal to $-e$ but im not 100% sure – Kinheadpump Jul 18 '19 at 07:29
  • Yes, that's exactly the conclusion that i came to myself, but again, i am looking to see if there is a way to solve this one, as I've said it is preparation for an exam. – Jon9 Jul 18 '19 at 07:37
  • What have you tried so far? – Kinheadpump Jul 18 '19 at 07:40
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    Select another answer – Claude Leibovici Jul 18 '19 at 07:40
  • I've tried some minor substitutions , but i have realised i cannot reach anywhere with it , even wolfram doesn't give a step-by-step solution . So is safe to say that you might have to deduce the answer . – Jon9 Jul 18 '19 at 07:48
  • This is quite a known integral around here, see: https://math.stackexchange.com/questions/2917524/calculate-int-limits-20-fracx-sqrtexx22dx and https://math.stackexchange.com/questions/2809087/integral-int-20-fracx-sqrtexx22dx?noredirect=1&lq=1 – Zacky Jul 18 '19 at 10:32

1 Answers1

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$$I =\int_{-2}^{0}\frac{xdx}{\sqrt{e^x+(x+2)^2}}=\int_{-2}^{0} \frac {xe^{-x/2} dx}{\sqrt{1+(x+2)^2e^{-x}}}=\int_{0}^{2} \frac{-2 dt}{\sqrt{1+t^2}}.$$ $$\Rightarrow I =-2 ~\mbox{arcsinh} t|_{0}^{2} =-2~\mbox{arcsinh} 2= -2 \ln(2+\sqrt{5})$$. Here we have used $t=(x+2)e^{-x/2}$.

Z Ahmed
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