5

I'm stuck with trying to integrate the following expression: $$ \int (\sqrt{x^2 + x}) dx $$

I have tried u-substitution where $u = \sqrt{x}$, but it didn't get far. How should I approach this?

Thanks ahead!

user321070
  • 193
  • The substitution to try would be $u = \sqrt{x^2+x}$. But then to find the inverse function, you solve a quadratic equation, and end up with Dr. MV's completion of the square anyway. – GEdgar Apr 18 '16 at 17:49

3 Answers3

7

HINT:

Note that

$$x^2+x=(x+1/2)^2-1/4$$

Then, use a standard trigonometric (or hyperbolic trigonometric) substitution.

Mark Viola
  • 179,405
  • i was about to write – Salihcyilmaz Apr 18 '16 at 17:33
  • Hyperbolic substitution, I'd say – Bernard Apr 18 '16 at 17:38
  • @Salihcyilmaz Perhaps you could post the complete solution. – Mark Viola Apr 18 '16 at 17:39
  • @Bernard Hyperbolic trigonometric substitution is subsumed in trigonometric substitution. But there is also nothing wrong with using the secant function here. – Mark Viola Apr 18 '16 at 17:40
  • @Dr. MV: OK, this is a matter of vocabulary… – Bernard Apr 18 '16 at 17:43
  • @Bernard Bernard, I've added a parenthetical reference to accommodate this vocabulary ambiguity. Thank you for the comment! Much appreciative. -Mark – Mark Viola Apr 18 '16 at 17:44
  • @Dr.MV I haven't learned substitution with trig functions, but I watched a few Khan's videos and gave it a shot, somehow got a negative in the square root... I can post the progress so far. – user321070 Apr 18 '16 at 18:01
  • 1
    @user321070 Substitute $(x+1/2)=\frac12 \sec(\theta)$ with $dx=\frac12 \sec(\theta)\tan(\theta),d\theta$. Then, recall $\sec^2(x)-1=\tan^2(x)$. The square root should then be devoid of any negative sign. ;-)) -Mark – Mark Viola Apr 18 '16 at 18:04
  • @Dr.MV Im studying for my Electromagnetics exam, too lazy to post solution for now, but you already gave a lot of hint with last comment:) – Salihcyilmaz Apr 18 '16 at 18:12
  • 1
    @Salihcyilmaz EM Theory is my specialty. My PhD Dissertation title was "An Integral-Operator Approach to the Electromagnetics of Integrated Optics." Let me know if I can help. -Mark – Mark Viola Apr 18 '16 at 18:19
  • @Dr.MV Took long time for me to edit in the progress, I just read your comment, I will take a closer look! – user321070 Apr 18 '16 at 18:30
  • @Dr.MV I managed to get to this point $ \frac{1}{4} \int( tan\theta sec\theta tan \theta ) d\theta $, not sure how to proceed? Should I take the integral at this point? – user321070 Apr 18 '16 at 19:18
  • @user321070 Write $\tan^2(\theta)\sec(\theta)=\sec^3(\theta)-\sec(\theta)$. The integral of the secant is a standard integral. For the integral of $\sec^3(\theta)$, integrate by parts with $u=\sec(\theta)$ and $v=\tan(\theta)$. – Mark Viola Apr 18 '16 at 19:36
  • @Dr.MV I just realized it, sorry I'm very bad with trig functions, I don't have all the formulas in my head, only the very very basic ones. Anyway, I tried to integrate $ \int sec^3 \theta $, and got kind of stuck in a loop at the $ \int u' v $ part. – user321070 Apr 18 '16 at 20:15
  • @Dr.MV First I get $ \int uv' = sec\theta tan\theta - \int sec\theta tan^2 \theta $, there if I integrate the $ \int u'v $ part, I get back $ tan\theta sec\theta - \int sec^3 \theta $. – user321070 Apr 18 '16 at 20:18
  • Note that the $\int u'v$ part looks just like the integral you were originally evaluating. HERE is a tutorial on the integral you are trying to evaluate. This should help. ;-)) -Mark – Mark Viola Apr 18 '16 at 20:18
  • @Dr.MV Thanks, I've too much more to learn. I think this question is probably out of difficulty for me at this time, I don't really think it will appear during exam (at least I hope). But after the last step, I need to substitute back $ x + \frac{1}{2} $ right? Am I supposed to take the $ arcsec $? – user321070 Apr 18 '16 at 20:31
  • You're quite welcome. My pleasure. You will need to substitute back $x+1/2=\frac12 \sec(\theta)$. – Mark Viola Apr 18 '16 at 20:37
1

Hint:

You can use this u-sub: $$ \sqrt{x^2+x}=u+x $$ $$ x^2+x=u^2+2ux+x^2$$ $$x-2ux=u^2 $$ $$x(1-2u)=u^2$$ $$x=\frac{u^2}{1-2u}$$ $$dx=\frac{-2u(u-1)}{(1-2u)^2}du$$

ultralegend5385
  • 4,245
ShishRobot
  • 319
0

For non-negative real $x$, you can use the following:

$$\begin{align*}x&=\sinh^2 z\\ \mathrm{d}x &= 2\sinh z \cosh z\,\mathrm{d}z\\\int \sqrt{x(x+1)}\, \mathrm{d}x &= 2 \int \sinh^2 z \cosh^2 z \, \mathrm{d}z\\&=\frac{1}{4}\int (\cosh 4z - 1) \mathrm{d}z \\ &\,\,\,\vdots \\ &= \frac{(2x+1)\sqrt{x^2+x}-\sinh^{-1}{\sqrt{x}}}{4}\end{align*}$$

Jack Tiger Lam
  • 4,394