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We have the following integral:

$$ \int_{2}^{7} \frac{x}{1-\sqrt{2+x}}\, dx $$

And this is my solution, which seems to be wrong, and I am failing to see where exactly I failed at:

We have $u=1-\sqrt{2+x}, x=u^2-2u-1, dx=-2\sqrt{2+x}\, du$, and we know that $x\geq -2$ and thus $u\leq 1$:

\begin{align} \int_{2}^{7} \frac{x}{1-\sqrt{2+x}}\, dx &=-2\int_{-1}^{-2} \frac{(u^2+2u-1)(u-1)}{u}\, du\\ &= -2 \left( \int_{-1}^{-2} u^2 d u + \int_{-1}^{-2} u\, du +\int_{-1}^{-2} -3\, du + \int_{-1}^{-2} \frac{1}{u}\, du \right) \\ &= -2\left[\frac{u^3}{3}+\frac{u^2}{2}-3u+\ln{|u|}\right]_{-1}^{-2}\approx -18 \end{align}

Can someone please help me pinpoint the issue?

DMcMor
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zareami10
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    $x=u^2-2u-1$ becomes in the integral $u^2+2u-1$. Another thing: $dx=2(u-1)du$, where did you get minus? – A.Γ. Jan 06 '21 at 18:20
  • Why should there be a negative sign outside your integral after substitution? And shouldn’t $+ 2u$ in the numerator be $- 2u$? – Ethan Mark Jan 06 '21 at 18:24
  • @EthanMark because $dx=-2\sqrt{2+x}, du$, or I am missing something? – zareami10 Jan 06 '21 at 18:33
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    @zareami10 I have posted an answer. Hopefully it is clearer for you there. Also, welcome to Math SE. It is good practice to upvote answers that have benefited you, as well accepting an answer that you feel is most suitable for you. – Ethan Mark Jan 06 '21 at 18:45
  • @EthanMark Thanks, and I will as soon as I have figured the problem out on my own. :) Currently struggling to understand the negative sign (albeit you're right about it). I mean $\frac{dx}{du}=\frac{1}{-2\sqrt{2+x}}$ so $dx=-2\sqrt{2+x}$, and then we have a negative sign. And I'm not sure what's wrong with that. Edit: Ah, I guess I'm missing that $u-1$ is always 0 and/or negative, so I should add another negative sign? – zareami10 Jan 06 '21 at 18:52
  • @EthanMark True, I have been missing that $\sqrt{(u-1)^2}$ should be $-(u-1)$ since it's either 0 or negative. – zareami10 Jan 06 '21 at 18:57
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    @zareami10 You are just making careless mistakes in your computation. So you have $\mathrm{d}x = -2\sqrt {2 + x}$ but $\sqrt {2 + x} = -(u - 1)$. Put both equations together and the negative sign vanishes. And as I mentioned in my answer, you are really just doing unnecessary work by writing $\mathrm{d}x$ in terms of $x$. – Ethan Mark Jan 06 '21 at 18:58
  • @EthanMark Albeit redundantly, I substituted $x$ directly, as in $dx=-2\sqrt{u^2-2u+1}$ so $dx=-2\sqrt{(u-1)^2}$, and that's why missing the negative out of the square root was the culprit, or so I think. – zareami10 Jan 06 '21 at 19:06
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    @zareami10 In any case, I hope I have cleared your doubts. – Ethan Mark Jan 06 '21 at 19:09

3 Answers3

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I would have calculated it like this:

Let $u=1-\sqrt{2+x} \Leftrightarrow x=(1-u)^2-2$ where $x\ge-2$. We then have $$ \mathrm dx=2(1-u)(-1)\,\mathrm du$$ so \begin{align*} \int_2^7\!\frac{x}{1-\sqrt{2+x}}\,\mathrm dx &=\int_{-1}^{-2}\!\frac{(1-u)^2-2}{u}\cdot2(1-u)(-1)\,\mathrm du \\&=2\int_{-2}^{-1}\!\frac{((1-u)^2-2)(1-u)}{u}\,\mathrm du \\&=2\int_{-2}^{-1}\!\frac{(1-2u+u^2-2)(1-u)}{u}\,\mathrm du \\&=2\int_{-2}^{-1}\!\frac{(u^2-2u-1)(1-u)}{u}\,\mathrm du \\&=2\int_{-2}^{-1}\!\frac{-u^3+3 u^2-u-1}{u}\,\mathrm du \\&=2\int_{-2}^{-1}\!\Bigl(-u^2+3u-1-\frac{1}{u}\Bigr)\,\mathrm du \\&=2\Bigl[-\frac13u^3+\frac32u^2-u-\ln(|u|)\Bigr]_{-2}^{-1} \\&=-\frac{47}{3}+2\ln(2). \end{align*}

mf67
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  • Correct me if I'm wrong, but I believe you forgot to multiply the $2$ into your $\frac {47} 3$? – Ethan Mark Jan 06 '21 at 18:42
  • @EthanMark I don't think so. Mathematica also returns -(47/3) + Log[4] as answer to the original integral. – mf67 Jan 06 '21 at 19:04
  • Indeed, the error was mine. I am writing this at 3 am. Brain probably needs some rest. Sorry about that! – Ethan Mark Jan 06 '21 at 19:14
  • No worries. I know the "3 am feeling" all too well myself. – mf67 Jan 06 '21 at 19:16
  • All the answers are pretty good, though since I can only accept one answer, I wanted to thank you (perhaps against the terms of using comments). – zareami10 Jan 06 '21 at 19:19
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The first thing I would like to point out is that since you want to do a substitution, it is really extra (unnecessary) work writing $dx$ in terms of $x$ (where the aim of substitution is to "get rid" of the initial variable and simplify the integral in the process).

With that out of the way, let $u = 1 - \sqrt {2 + x}$, then $x = u^2 - 2u - 1$ and $\mathrm{d}x = 2u - 2$.

Now, observe that, $\forall\ n \in \mathbb {R}$, $\sqrt n \geq 0$, so we note that $u \leq 1$.

\begin{align} \int_{2}^{7} \frac x {1 - \sqrt {2 + x}}\ \mathrm {d}x & = 2\int_{-1}^{-2} \frac {(u^2 - 2u - 1)(u - 1)} u\ \mathrm {d}u \\[5 mm] & = 2\int_{-1}^{-2} \frac {u^3 - 3u^2 + u + 1} u\ \mathrm {d}u \\[5 mm] & = 2\left[\frac {u^3} 3 - \frac {3u^2} 2 + u + \ln|u|\right]^{-2}_{-1} \\[5 mm] & = 2\ln 2 -\frac {47} 3 \end{align}

Ethan Mark
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exactly, when you have $u=1-\sqrt{2+x}\ => x=u^2-2u-1$

Then $du=2(u-1)dx$

Plugging it into your formula we get

$=2\int\ \frac{(u^2-2u-1)(u-1)}{u}$

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