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My question is: How can I integrate $\sqrt{1+\frac{1}{3x}} \, dx$ ? (I don't mind about negative Xs).

I'm aware that there's a theorem which you can use that integrates the function using the points where the function can't be solved or something like that(improper integrals?) but I haven't learned anything but parts and substitution yet. I know that this integral can be solved using just those and that's what I am searching for =).

Unfortunately, this is no homework (and my calculus teacher has no idea how to solve this... I asked =( ). I just saw this integral by accident. I've been struggling with it for like a week now. I just want the answer and preferably the steps. Possible starting points:


The best approach I could find is:

$$u^2=3x \implies 2u\ du = 3\ dx$$

$$\frac{2}{3}\int \sqrt{ 1+\frac{1}{u^2} } \, \, u\ du$$

$$\frac{2}{3}\int \sqrt{ u^2+1 } \, \, du$$

Now I know that I can use sinh but I have no clue how.

That was one approach that a guy told me. If you don't like it you could start with the more traditional way:

$$u=3x$$

$$ \frac{1}{3} \int\sqrt{1+\frac{1}{u}}\,du.$$ Now $$w^2=1+\dfrac{1}{u}$$ $$2w\,dw=-\frac{du}{u^2}=-(w^2-1)^2 \,du.$$ $$\frac{1}{3}\int-\frac{2w^2\,dw}{(w^2-1)^2}.$$

After that I've been told that you could integrate by parts but it's just too hard for me. I simply get nowhere. I won't type everything cause it's a waste of time. If anyone wants me to type in more work I'd be happy to if it helps in any way.

I'm in a stage where I just want to see how this super-complex (at least for me) integral is solved. I've used wolframalpha but it's nowhere near a human approach. So well... thanks a lot for any help guys! And sorry for the long post! =)!

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  • $\begingroup$ To call this primitive weird is uninformative and... well, weird. $\endgroup$ – Did Oct 29 '12 at 17:52
  • $\begingroup$ @did well it's weird for me.... if you want I can change the title.... $\endgroup$ – Gaspa79 Oct 29 '12 at 18:10
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Method 1: Let's try integration by parts \begin{eqnarray*} \int\sqrt{u^2+1}\,du&=&\int1\cdot\sqrt{u^2+1}\,du=u\sqrt{u^2+1}-\int u\frac{2u}{2\sqrt{u^2+1}}\,du=\\ &=&u\sqrt{u^2+1}-\int \frac{u^2+1-1}{\sqrt{u^2+1}}\,du= u\sqrt{u^2+1}-\int\left(\sqrt{u^2+1}-\frac{1}{\sqrt{u^2+1}}\right)\,du=\\ &=&u\sqrt{u^2+1}-\underbrace{\int\sqrt{u^2+1}\,du}_{\text{same}}+\int\frac{1}{\sqrt{u^2+1}}\,du. \end{eqnarray*} The unknown integral can be solved now as $$ 2\int\sqrt{u^2+1}\,du=u\sqrt{u^2+1}+\int\frac{1}{\sqrt{u^2+1}}\,du $$ which gives you the answer if you know how to calculate the last integral (quite standard). If you don't then you can try

Method 2: The standard substitution for "square root of the square plus constant" $$ t=u+\sqrt{u^2+1} $$ that gives after some algebra $$ u=\frac{t^2-1}{2t}=\frac{t}{2}-\frac{1}{2t},\quad du=\frac{t^2+1}{2t^2}\,dt,\quad \sqrt{u^2+1}=t-u=\frac{t^2+1}{2t} $$ $$ \int\sqrt{u^2+1}\,du=\int\frac{(t^2+1)^2}{4t^3}\,dt=\int\frac{t^4+2t^2+1}{4t^3}\,dt=\frac{1}{4}\int\left(t+\frac{2}{t}+\frac{1}{t^3}\right)\,dt $$ which I believe you can manage yourself.

P.S. When doing the final substitution back to $u$ it is beneficial to note that $$ \frac{1}{t}=\frac{1}{\sqrt{u^2+1}+u}=\frac{1}{\sqrt{u^2+1}+u}\cdot\frac{\sqrt{u^2+1}-u}{\sqrt{u^2+1}-u}=\frac{\sqrt{u^2+1}-u}{u^2+1-u^2}=\sqrt{u^2+1}-u. $$

P.P.S. Can you now calculate the last integral in the method 1 by the method 2?

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This is a nice candidate for the “integral of the inverse” technique.
The idea is to use the integration by parts formula in a weird way: $$\int {y dx} = xy - \int{x dy}$$ Let $ y= \sqrt{1+ \frac{1}{3x} } $.
So $x=\frac{1}{3} \frac{1}{y^2-1} $

Plugging this into the integration by parts formula: $$ \int{\sqrt{1+\frac{1}{3x}}dx} = x\sqrt{1+\frac{1}{3x}} - \frac{1}{3}\int {\frac{dy}{y^2-1}} $$ which is much easier to integrate. Afterwards, substitute $ y= \sqrt{1+ \frac{1}{3x} } $ for any remaining y’s.

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\begin{equation} \int\sqrt{1+\frac{1}{3x}}\ dx \end{equation} Let \begin{align*} x&=\frac{\cot^2\theta}{3},\\ \frac{\rm{d}x}{\rm{d}\theta}&=\frac{2}{3}(\cot\theta)(-\csc^2\theta) \end{align*} Use the above substitution in (1) \begin{equation}\begin{split} -\frac{2}{3}\int\sqrt{1+\tan^2\theta}\cot\theta\csc^2\theta \ d\theta &=-\frac{2}{3}\int\sec\theta\cot\theta\csc^2\theta \ d\theta\\ &=-\frac{2}{3}\int\frac{\cot\theta}{\cos\theta\sin^2\theta}\ \rm{d}\theta\\ &=-\frac{2}{3}\int\csc^3\theta \ d\theta \end{split}\end{equation} From here, we need integration by parts \begin{align*} u&=\csc\theta,\quad &dv=\csc^2\theta \ d\theta,\\ \rm{d}u&=-\cot\theta\csc\theta \quad &v=-\cot\theta \end{align*} We now have \begin{equation*} \begin{split} -\frac{2}{3}\int\csc^3\theta \ \rm{d}\theta&=-\frac{2}{3}\left[-\csc\theta\cot\theta-\int\cot^2\theta\csc\theta \ d\theta\right]\\ &=-\frac{2}{3}\left[-\csc\theta\cot\theta-\int(\csc^2\theta-1)\csc\theta \ d\theta\right]\\ &=-\frac{2}{3}\left[-\csc\theta\cot\theta-\int\csc^3\theta\ d\theta +\int\csc\theta)\ \rm{d}\theta\right]\\ &=-\frac{2}{3}\left[-\csc\theta\cot\theta-\int\csc^3\theta\ d\theta +\int\csc\theta\left(\frac{\csc\theta-\cot\theta}{\csc\theta-\cot\theta}\right)\ d\theta\right]\\ &=-\frac{2}{3}\left[-\csc\theta\cot\theta-\int\csc^3\theta\ d\theta +\int\frac{\csc^2\theta-\csc\theta\cot\theta}{\csc\theta-\cot\theta}\ d\theta\right] \end{split} \end{equation*} Now, in the rightmost integral we make the substitution $$u=\csc\theta-\cot\theta, \quad\rm{d}u=(\csc^2\theta-\csc\theta\cot\theta)\ d\theta$$ and we end up with \begin{equation*} -\frac{2}{3}\int\csc^3\theta\ d\theta=\frac{2}{3}\csc\theta\cot\theta+\frac{2}{3}\int\csc^3\theta\ \rm{d}\theta-\frac{2}{3}\ln(\csc\theta-\cot\theta) \end{equation*} Finally collecting similar terms we get \begin{equation*} \begin{split} -\frac{4}{3}\int\csc^3\theta\ d\theta&=\frac{2}{3}\csc\theta\cot\theta-\frac{2}{3}\ln(\csc\theta-\cot\theta),\\ \int\csc^3\theta\ d\theta&=-\frac{1}{2}\csc\theta\cot\theta+\frac{1}{2}\ln(\csc\theta-\cot\theta), \end{split} \end{equation*} So the final result is $$\int\sqrt{1+\frac{1}{3x}}\ \rm{d}x=\frac{1}{2}\left[\ln(\csc\theta-\cot\theta)-\csc\theta\cot\theta\right]$$

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For your first approach, $\int \sqrt{u^2+1} du$, you could try trig substitution, i.e. you may let $u = \tan(\theta)$, and use the identity $\tan^2(\theta)+1 = \sec^2(\theta)$, the integral after the substitution is $\int \sec^3 \theta d\theta$, to do this one you can find it on wikipedia: http://en.wikipedia.org/wiki/Integral_of_secant_cubed

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  • $\begingroup$ Okay after 20 minutes I realized that I am in a dead end again. After solving the integrals I end with different trigonometric functions (I used wikipedia as you said.. yay!). What should I do to replace them in form of u? Should I find the trigonometric identity between the functions I end with and the tangent? Thanks a lot! =) $\endgroup$ – Gaspa79 Oct 29 '12 at 19:02
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    $\begingroup$ I believe you find something like $\int \sec^3 xdx=\frac{1}{2}\sec x \tan x+ \frac{1}{2} \log|\sec x + \tan x|$ as in the Wikipedia article, now you know that $u = \tan \theta$, so the matter is to figure out how to represent $\sec(\theta)$ in terms of $u$, but again by the identity we have $1+u^2=1+\tan^2 (\theta) = \sec^2 \theta $ $\endgroup$ – TTY Oct 29 '12 at 21:09
  • $\begingroup$ This was the way to go. By the time I read your comment I had already "solved it". I watched a khanacademy video on trigonometric substitutions and I understood how to do it. I honestly can't believe that I (well.. actually YOU) solved it. THANKS A LOT!!! $\endgroup$ – Gaspa79 Oct 29 '12 at 21:50
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I think the following substitution is easier to work with:

$$u^2=1+\frac{1}{3x}\Longrightarrow 2u\,du=-\frac{dx}{3x^2}\Longrightarrow dx=-6u\frac{1}{9(1-u^2)^2}\,du\Longrightarrow$$

$$\Longrightarrow \int\sqrt{1+\frac{1}{3x}}dx=-\frac{2}{3}\int\frac{u^2}{(1-u^2)^2}du$$

which is already a rational integral (partial fractions and etc.).

Added $\;\;\;$ Partial fractions:

$$\frac{u^2}{(u^2-1)^2}=\frac{A}{u-1}+\frac{B}{(u-1)^2}+\frac{C}{(u+1)}+\frac{D}{(u+1)^2}\Longrightarrow$$

$$u^2=A(u-1)(u+1)^2+B(u+1)^2+C(u-1)^2(u+1)+D(u-1)^2$$

In the last polynomial identity assign values to u (recommended: $\,u=0\,,\,\pm1\,$) and compare powers of the variable (say, of $\,u^3\,$) in order to get the RHS coefficients. If I didn't make a mistake ( and I wouldn't waige on this!), one gets

$$A=B=D=\frac{1}{4}=-C$$

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  • $\begingroup$ Wait... is that even possible? I'm at work so I'm doing calculations on my head... Wow.. just... wow. So how can I go on from here? Should I integrate by parts? (one part $u^2$ another part $(1-u^2)^2$). Or should I try to find some substitution after cancelling the powers? Thanks a lot I'll try as soon as I go home!! =) $\endgroup$ – Gaspa79 Oct 29 '12 at 17:59
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    $\begingroup$ Read my added stuff in my answer $\endgroup$ – DonAntonio Oct 29 '12 at 18:07
  • $\begingroup$ WTF? I'm guessing I'll need to learn partial fractions then. I'll read about partial fractions and watch a video or two. After that can I come back to you? =) $\endgroup$ – Gaspa79 Oct 29 '12 at 18:17
  • $\begingroup$ Ya me parecia conocido tu nick! Sos el mexicano que me ayudo cuando estaba aprendiendo limites! Jajaja muchas gracias =)!! $\endgroup$ – Gaspa79 Oct 29 '12 at 18:18
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    $\begingroup$ ¿Tú sabes (o en lunfardo: ¿vos sabés?) a cuántos miles de personas he encontrado a lo largo de los años en estos sitios? Igual me da gusto el poder haberte ayudado $\endgroup$ – DonAntonio Oct 29 '12 at 18:36
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Another strategy could be

$$\int\sqrt{1+\frac{1}{3x}}dx=\int\sqrt{\frac{1}{3x}\left(1+3x\right)}dx=\frac{2}{\sqrt3}\int\frac{1}{2\sqrt{x}}\left(\sqrt{1+3x}\right)dx$$

Now let $\sqrt{x}=y$, hence $\frac{1}{2\sqrt{x}}dx=dy$

$$\frac{2}{\sqrt3}\int\frac{1}{2\sqrt{x}}\left(\sqrt{1+3x}\right)dx=\frac{2}{\sqrt3}\int\sqrt{1+3y^2}dy$$

Now let $\sqrt{3}y=\sinh z$, hence $dy=\frac{\cosh z}{\sqrt3}dz$

$$\frac{2}{\sqrt3}\int\sqrt{1+3y^2}dy=\frac{2}{3}\int(\cosh z)^2dz=\frac{1}{6}\int(e^z-e^{-z})^2dz=\frac{1}{12}e^{2z}-\frac{1}{12}e^{-2z}+\frac{1}{3}z+c$$

Finally we have

$$\int\sqrt{1+\frac{1}{3x}}dx=\frac{1}{12}e^{2arcsinh(\sqrt{3x})}-\frac{1}{12}e^{-2arcsinh(\sqrt{3x})}+\frac{1}{3}arcsinh(\sqrt{3x})+c$$

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