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$$\int\frac{\sqrt{a+x}}{\sqrt{a}+\sqrt{x}}\, dx$$

A few days ago I asked a similar (looking) question where all the pluses here were minuses. It could be much more easily manipulated than this one using difference of $2$ squares. The other method used to solve the previous problem is replacing $x$ with $\arccos^2x$. Out of curiosity I have attempted to use the latter method to solve this integral, and made a little progress with it.

Just to make sure this integral has a solution, Wolframalpha found a solution that was not too bad.

Replacing $x$ with $\arctan^2t$, the integral becomes

$$\int\frac{2a \sec^3 t \tan t}{1+\tan t}\, dt$$

From here I tried to perfrom integration by parts, where $2a$ is taken outside of the integral, and numberator split into $sec x\tan x$ times $\sec^2x$, as $u$ and $v'$ respectively. $v$ can easily be found but $u'$ is messy. Integration by parts needed to be done again, also, which would have ended up really messy.

So I tried to convert everything into sines and cosines, then letting $u$ equal $\cos t$, and $du$ equal to $-\sin tdt$.

$$\int\frac{2a\sin t}{\cos^3 t \cos t +\sin t}\, dt$$

$$\int\frac{-2au}{u^3 u+\sqrt{1-u^2}}\, dt$$

However that doesn't look very pretty either. I don't think it is possible to perform Partial Fraction Decomposition on it.

What are the right paths to take to solve this integral? Are there any 'tricks of the trade' I have missed along the way?

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  • $\begingroup$ This was solve here before and not by that hideous trigonometric substitution. $\endgroup$
    – OR.
    Commented Jul 16, 2013 at 23:55
  • $\begingroup$ I think you're misusing backslash. $\endgroup$
    – Kaster
    Commented Jul 16, 2013 at 23:58
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    $\begingroup$ I do not know where the $\tan t$ on top comes from. If you really want to imitate the trigonometric substitution (which is not necessarily a good idea) then $x=a\sinh^2 t$ looks better. $\endgroup$ Commented Jul 17, 2013 at 0:05
  • $\begingroup$ Kaster: please give advice; i'm new. RGB: can you send me the link to the solution or what was 'solved here before'? thanks $\endgroup$
    – Y-dog
    Commented Jul 17, 2013 at 0:06
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    $\begingroup$ If you asked it, it is in your profile. (math.stackexchange.com/questions/443824/…) $\endgroup$
    – OR.
    Commented Jul 17, 2013 at 0:10

5 Answers 5

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We multiply by $1 = \frac{\sqrt{a} - \sqrt{x}}{\sqrt{a} - \sqrt{x}}$: $$\int\frac{\sqrt{a+x}}{\sqrt{a}+\sqrt{x}}dx = \int\frac{\sqrt{a^2+ax} - \sqrt{ax + x^2}}{a - x}dx$$ Now we can separate the integral into two parts. For the left part: $$\int\frac{\sqrt{a^2+ax}}{a - x}dx = \int\frac{\sqrt{y}}{2a^2-y}dy = \frac{\sqrt{2}}{a}\int\frac{z^2}{a\sqrt{2}+z}+\frac{z^2}{a\sqrt{2}-z}dz$$ After substitutions $y = a^2 + ax$ and $z^2 = y$. This integral evaluates easily to $$4z-8+2\sqrt{2}a(\log(z+a\sqrt{2}) - \log(z-a\sqrt{2})) =$$ $$=4\sqrt{a^2+ax}-8+2\sqrt{2}a(\log(\sqrt{a^2+ax}+a\sqrt{2}) - \log(\sqrt{a^2+ax}-a\sqrt{2}))$$ I haven't tried yet to evaluate the other integral.

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    $\begingroup$ you made the mistake $(\sqrt{a}+\sqrt{x})(\sqrt{a}-\sqrt{x})\neq a+x$ $\endgroup$
    – Norbert
    Commented Jul 19, 2013 at 20:28
  • $\begingroup$ @Norbert Oh, hell! You're right of course. $\endgroup$ Commented Jul 19, 2013 at 20:33
  • $\begingroup$ This half worked out quite neatly. The other half looks much nastier. Any ideas as to nudge out the other half? I don't think a similar substitution would work. Your method seems to be the neatest so far! $\endgroup$
    – Y-dog
    Commented Jul 21, 2013 at 4:50
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    $\begingroup$ @YunFeiOuYang Thanks :) For the second part, I have tried various techniques, but got nowhere. The best I have it the substitution $y = x+a$, which gives $\int \frac{\sqrt{ax + x^2}}{a-x}dx = \int \sqrt{1-\frac{a}{y}}dy$, but then I don't know how to go on. $\endgroup$ Commented Jul 21, 2013 at 17:56
  • $\begingroup$ @Daniel: Your solution looks a little like the answer by Wolframalpha, especially the part of 2sqrt(2)a. (see #solution in original question). It is really surprising how something so ordinary looking can be so hard! :) $\endgroup$
    – Y-dog
    Commented Jul 22, 2013 at 12:35
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One way to approach this problem is to use the following substitutions:

$b^2=a$ and $x=b^2u^2$ and $u=\frac{2v}{1-v^2}\implies\sqrt{1+u^2}=\frac{1+v^2}{1-v^2}\text{ and }\mathrm{d}u=2\frac{1+v^2}{(1-v^2)^2}\mathrm{d}v$

For example $$ \begin{align} \int\frac{\sqrt{a+x}}{\sqrt{a}+\sqrt{x}}\,\mathrm{d}x &=\int\frac{\sqrt{b^2+b^2u^2}}{b+bu}2b^2u\,\mathrm{d}u\\ &=2b^2\int\frac{\sqrt{1+u^2}}{1+u}u\,\mathrm{d}u\\ &=8b^2\int\frac{v(1+v^2)^2}{(1+2v-v^2)(1-v^2)^3}\,\mathrm{d}v \end{align} $$ Now, at least, the problem can be handled with partial fractions. $$ \begin{align} \frac{8v(1+v^2)^2}{(1+2v-v^2)(1-v^2)^3} &=\frac2{(v+1)^3}+\frac1{(v+1)^2}+\frac3{(v+1)}\\ &-\frac2{(v-1)^3}-\frac3{(v-1)^2}-\frac3{(v-1)}\\ &+\frac{2\sqrt2}{(v-1-\sqrt2)}-\frac{2\sqrt2}{(v-1+\sqrt2)} \end{align} $$

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  • $\begingroup$ This method yields ok results. Partial fraction decomposition : wolframalpha.com/widgets/… $\endgroup$
    – Y-dog
    Commented Jul 20, 2013 at 9:34
  • $\begingroup$ It will take long time to handle the partial fractions. $\endgroup$
    – xpaul
    Commented Jul 20, 2013 at 13:02
  • $\begingroup$ Substitution back will be a problem because v can only be expressed as two quadratic roots of u. Is there a way around this? $\endgroup$
    – Y-dog
    Commented Jul 23, 2013 at 8:12
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Take $x=az$ and $z=s^2$, where $s\geq 0$. Then

\begin{aligned} \int\frac{\sqrt{a+x}}{\sqrt{a}+\sqrt{x}}\, dx &=a \int\frac{\sqrt{1+z}}{1+\sqrt{z}}\, dz \\ &=2a \int \sqrt{1+s^2}\frac{s}{1+s}\, ds \\ &=2a \int \sqrt{1+s^2}\,ds-2a \int \frac{\sqrt{1+s^2}}{1+s}\, ds \\ &=2aI_1-2aI_2. \end{aligned}

Here (using the standard substitution $s=\sinh(t)$) $$ I_1=\frac{1}{2}s\sqrt{1+s^2}+\frac{1}{2}\operatorname{arcsinh}(s)+C. $$ Now consider $I_2$.

In the integrand $\sqrt{1+s^2}/(1+s)$, substitute $s=\tan(u),\,(0\leq u<\pi/2)$ and $ds=du \sec^2(u)$. Then $\sqrt{1+s^2}=\sqrt{1+\tan^2(u)}=\sec(u)$ and $u=\tan^{-1}(s)$, $$ I_2=\int\frac{\sec^3(u)}{1+\tan(u)}\,du. $$ In the integrand $\sec^3(u)/(1+\tan(u))$, substitute (the standard) $p=\tan(u/2),\,(0\leq p<1)$ and $dp=(1/2)du\sec^2(u/2)$. Then using this substitution $\sin(u)=(2 p)/(1+p^2)$, $\cos(u)=(1-p^2)/(1+p^2)$ and $du=(2 dp)/(1+p^2)$ we obtain $$ I_2=\int 2 \frac{(1+p^2)^2}{(1-p^2)^3 (1+\frac{2 p}{1-p^2})} dp=\int-\frac{2(1+p^2)^2}{(-1+p^2-2p)(-1+p^2)^2}\,dp. $$ Here (using computer) $$ \begin{aligned} -\frac{2(1+p^2)^2}{(-1+p^2-2p)(-1+p^2)^2} &=\frac{1}{(p-1)^2}-\frac{4}{-1+p^2-2p} \\ &=-\frac{1}{p+1}+\frac{1}{p-1}-\frac{1}{(p+1)^2}. \end{aligned} $$ These 5 integrals can be calculated easily. Now you should make back substitutions and applying a lot of trigonometric identities to obtain the final answer. I omit the details because you asked about methods ('tricks') of calculating integrals.

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  1. Avoid trigonometric substitutions, they only work in specially designed examples, i.e. if it is your only tool you will easy find exercises in which they are a pain.
  2. Rationalize the denominator, by using difference of squares as you have seen before.
  3. Rewrite as a sum of integrals of the form $\int R(\sqrt{\frac{ax+b}{cx+d}})\text{d}x$, where $R$ are rational functions.
  4. Apply the substitution $y=\frac{ax+b}{cx+d}$, which always work for all of these.

Addendum: For that integral that you got at the end you can use Euler's substitutions. This turns that last integrand into a rational function and from there partial fraction decomposition finishes it. Still, the trigonometric substitution was a long path to get to the rational function.

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With $x=a \sinh^2 t$ \begin{align} &\frac1a\int\frac{\sqrt{a+x}}{\sqrt{a}+\sqrt{x}}\, dx = 2\int \frac {\sinh t \cosh^2 t}{1+\sinh t}dt\\ = &\ 2\int \cosh^2t-\sinh t+1-\frac2{1+\sinh t}\ dt\\ =&\ \frac12\sinh2t -2\cosh t+3t -2\sqrt2 \tanh^{-1}\frac{\sinh t-1}{\sqrt2\cosh t}\\ =&\ \sqrt{\frac xa +1}\left(\sqrt{\frac xa} -2\right) +3\sinh^{-1}\sqrt {\frac xa} -2\sqrt2 \tanh^{-1}\frac{\sqrt{x}-\sqrt a}{\sqrt2\sqrt{x+a}} \end{align}

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