Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
4  
Try just $u=2+x$. –  David Mitra May 9 '13 at 1:46
2  
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. –  User58220 May 9 '13 at 2:01
    
@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.) –  le gâteau au fromage Aug 1 at 6:51

3 Answers 3

up vote 2 down vote accepted

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 $$\int (u^2-2)(u)(2u)\,du=\int (2u^4-4u^2)\,du=\frac{2u^5}{5}-\frac{4u^3}{3}+C.$$

share|improve this answer
    
Clear and concise, thank you! –  seeker May 9 '13 at 2:55

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

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

which can now be integrated easily.

share|improve this answer
    
Nice advice, thanks! –  seeker May 9 '13 at 1:55

I think it is simpler by parts:

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

so

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.