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Calculate $$\int\limits_{-2}^{0} \frac{x}{\sqrt{e^x+(x+2)^2}}dx$$

First I tried the substitution $t=x+2$ and obtained $$\int\limits_{0}^{2} \frac{t-2}{\sqrt{e^{t-2}+t^2}}dt$$ and than I thought to write it as $$\int\limits_{0}^{2} (t-2)\frac{1}{\sqrt{e^{t-2}+t^2}}dt$$ and use the fact that $$2(\sqrt{e^{t-2}+t^2})'=\frac{1}{\sqrt{e^{t-2}+t^2}} \cdot(e^{t-2}+2t)$$ Using integration by parts we get that we have to calculate (excluding some terms we know) $$\int\limits_{0}^{2} \sqrt{e^{t-2}+t^2}\cdot \frac{6e^{t-2}-2te^{t-2}+8}{(e^{t-2}+2t)^2}dt$$ which is uglier then the initial problem. Do you have any idea how to solve the problem?

razvanelda
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  • Wolframalpha's calculation: http://www.wolframalpha.com/input/?i=integrate+x%2Fsqrt(e%5Ex%2B(x%2B2)%5E2)+from+-2+to+0 However, is it possible to find the closed form of the given integral? – choco_addicted Sep 15 '18 at 05:52
  • Did you find this integral in a book? Perhaps previous problems or the topic of the chapter can give some clue. Note also that it is a definite integral, and that finding a primitive may not be the best way to compute it. – ajotatxe Sep 15 '18 at 05:54
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    It was also answered here: https://math.stackexchange.com/q/2809087/515527 and here https://math.stackexchange.com/q/2699140/515527 – Zacky Sep 15 '18 at 07:42
  • @Zacky oh I had no idea :)) thanks – razvanelda Sep 15 '18 at 07:49
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    @Zacky Thank you, i will use it next time – razvanelda Sep 15 '18 at 09:16
  • @Zacky I agree with you about the duplicate comment but with your comment it's not possible to figure out which one is the original one which is why that "possible duplicate" comment is there to show which is original and to indicate that the post has been flagged. Also, the comment was automatically added after I flagged it. I also flagged another link that you mentioned as "possible duplicate". My comment would automatically be deleted after this post is closed. That is how it works. – paulplusx Sep 15 '18 at 11:28
  • @Zacky So from next time if you find a duplicate just flag it as duplicate and a comment of "possible duplicate" will be added for others as an indicator that the post has been flagged. No need to manually put the link in the comment. – paulplusx Sep 15 '18 at 11:35

2 Answers2

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Hint: Let $y = (x+2)e^{-x/2}.$ You should get $I=-2\sinh^{-1}{2}$.

$\displaystyle y = (x+2)e^{-x/2} \implies dy = -\frac{x}{2}e^{-x/2}\,dx$ and $\displaystyle \frac{x\,dx}{\sqrt{e^x+(x+2)^2}} = \frac{xe^{-x/2}\,dx}{\sqrt{1+(x+2)^2e^{-x}}} = \frac{-2\,dy}{\sqrt{1+y^2}}.$

Therefore we have $\displaystyle I = \int_{-2}^{0}\frac{x\,dx}{\sqrt{e^x+(x+2)^2}} = -2\int_{0}^{2}\frac{dy}{\sqrt{1+y^2}} = -2\sinh^{-1}(2)$, as claimed.

Migos
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Long method One possible method is the following. Using \begin{align} \sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{(-1)^n \, x^n}{4^n} &= \frac{1}{\sqrt{1 + x}} \\ \gamma(s,x) &= \int_{0}^{x} e^{-t} \, t^{s-1} \, dt \end{align} then: \begin{align} I &= \int_{-2}^{0} \frac{x \, dx}{\sqrt{e^{x} + (x+2)^{2}}} \\ &= \int_{-2}^{0} \frac{x \, e^{-x/2} \, dx}{\sqrt{1 + \left(\frac{(x+2)^2}{e^{x}}\right)}} \\ &= \sum_{n} \binom{2n}{n} \frac{(-1)^{n}}{4^{n}} \, \int_{-2}^{0} e^{-(n+1/2) x} \, x \, (x+2)^{2n} \, dx \\ &= \sum_{n} \binom{2n}{n} \frac{(-1)^{n}}{4^n} \, e^{2n+1} \, \int_{0}^{2} e^{-(n+1/2) t} \, (t-2) \, t^{2n} \, dt \hspace{5mm} t \to x + 2 \\ &= \sum_{n} \binom{2n}{n} \, \frac{(-1)^{n}}{4^{n}} \, e^{2n+1} \, \left[ \left(\frac{2}{2n+1}\right)^{2n+2} \, \gamma(2n+2, 2n+1) - 2 \, \left(\frac{2}{2n+1}\right)^{2n+1} \, \gamma(2n+1, 2n+1) \right] \\ &= 4 \, \sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{(-1)^{n} \, e^{2n+1}}{(2n+1)^{2n+2}} \, \left[ \gamma(2n+2, 2n+1) - (2n+1) \, \gamma(2n+1, 2n+1) \right] \\ &= - 4 \, \sum_{n} \binom{2n}{n} \, \frac{(-1)^{n} \, e^{2n+1}}{(2n+1)^{2n+2}} \, (2n+1)^{2n+1} \, e^{-(2n+1)} \\ &= -4 \, \sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{(-1)^{n}}{(2n+1)} \\ &= - \, \sum_{n} \binom{2n}{n} \frac{(-1)^{n}}{4^{n}} \, \int_{0}^{2} y^{2n} \, dy \\ &= -2 \, \int_{0}^{2} \frac{dy}{\sqrt{1+y^2}} \\ &= -2 \, \sinh^{-1}(2). \end{align}

Short method

\begin{align} I &= \int_{-2}^{0} \frac{x \, dx}{\sqrt{e^{x} + (x+2)^{2}}} \\ &= -2 \int_{-2}^{0} \frac{- (x/2) \, e^{-x/2} \, dx}{\sqrt{1 + ((x+2) \, e^{-x/2})^2}} \\ \end{align} Let $u = (x+2) \, e^{-x/2}$, $du = -(x/2) \, e^{-x/2} \, dx$, then \begin{align} I &= -2 \int_{-2}^{0} \frac{du}{\sqrt{1+u^2}} = -2 \left[\sinh^{-1}(u)\right]_{-2}^{0} \\ &= -2 \, \sinh^{-1}(2). \end{align}

It is of interest to note that $$\int_{-2}^{\infty} \frac{x \, dx}{\sqrt{e^{x} + (x+2)^{2}}} = 0.$$

Leucippus
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