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My attempt at the question:

$\eqalign{ & \int {x\sqrt {2 + x} {\rm{ }}dx} \cr & {u^2} = 2 + x \cr & 2u{{du} \over {dx}} = 1 \cr & {{du} \over {dx}} = {1 \over {2u}} \cr & u = \sqrt {2 + x} \cr & x = {u^2} - 2 \cr & so: \cr & \int {x\sqrt {2 + x} {\rm{ }}dx} = \int {x\sqrt {2 + x} } {\rm{ }}{{dx} \over {du}}du \cr & = \int {x\sqrt {2 + x} } {\rm{ }} \times 2\sqrt {2 + x} du \cr & = \int {2x} (2 + x)du \cr & = \int {4x} + 2{x^2}du \cr & = \int {4({u^2} - 2)} + 2{({u^2} - 2)^2}du \cr & = \int {4{u^2} - 8} + 2({u^4} - 4{u^2} + 4)du \cr & = \int {2{u^4} - 4{u^2}} du \cr & = {2 \over 5}{u^5} - {4 \over 3}{u^3} + C \cr & = {2 \over 5}{(\sqrt {2 + x} )^5} - {4 \over 3}{(\sqrt {2 + x} )^3} + C \cr & = {2 \over 5}{(2 + x)^{{5 \over 2}}} - {4 \over 3}{(2 + x)^{{3 \over 2}}} + C \cr} $


A few questions I have:

Given ${u^2} = 2 + x$, $u = \pm \sqrt {2 + x} $, so why is it that we only take the principal square root and not the negative one for substitution?

The second question I have is a general one; is there an easier way of finding the integral? Have I done things in a manner that isn't overly longwinded? If so please suggest ways that would allow me to reach an answer quicker.

I'm on shakey grounds with integration at the moment so I was wondering if I could integrate this part of my working out without expanding out:

$ = \int {4({u^2} - 2)} + 2{({u^2} - 2)^2}du$

Thank you for all your help!

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    $\begingroup$ Try just $u=2+x$. $\endgroup$ May 9, 2013 at 1:46
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    $\begingroup$ Amplifying on David Mitra's comment: an integral with a simple linear expression to a simple power, times a complicated linear to a complicated power, will very often benefit by his simple switch of roles. $\endgroup$
    – DJohnM
    May 9, 2013 at 2:01
  • $\begingroup$ @Assad I just want to comment that I personally liked your work presentation on evaluating this integral via the substitution rule. It's just that every conceivable detail is listed; not a single step is skipped. (Most mathematicians like to skip steps, for brevity purposes I think.) $\endgroup$
    – Cookie
    Aug 1, 2014 at 6:51

3 Answers 3

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If $u=2+x \implies dx=du$ then the integral becomes

$$\int (u-2)\sqrt u\,du$$

which can now be integrated easily.

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  • $\begingroup$ Nice advice, thanks! $\endgroup$
    – seeker
    May 9, 2013 at 1:55
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Let's use your substitution, without the unnecessary manipulations.

Let $u^2=x+2$. Then $2u\,du=dx$ and $x=u^2-2$. Substitute, getting rid of all $x$ all at once. We get $$\int (u^2-2)(u)(2u)\,du=\int (2u^4-4u^2)\,du=\frac{2u^5}{5}-\frac{4u^3}{3}+C.$$

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  • $\begingroup$ Clear and concise, thank you! $\endgroup$
    – seeker
    May 9, 2013 at 2:55
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I think it is simpler by parts:

$$u=x\;,\;\;u'=1\\v'=\sqrt{x+2}\;,\;\;v=\frac23(x+2)^{3/2}$$

so

$$\int x\sqrt{x+2}\,dx=\frac23x(x+2)^{3/2}-\frac23\int(x+2)^{3/2}dx=\frac23x(x+2)^{3/2}-\frac4{15}(x+2)^{5/2}+C=$$

$$=\frac2{15}(x+2)^{3/2}\left(3x-4\right)+C$$

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