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

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?

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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


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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

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. $$ \begin{align} \frac{8v(1+v^2)^2}{(1+2v-v^2)(1-v^2)^3} &=\frac2{(v+1)^3}+\frac1{(v+1)^2}+\frac3{(v+1)}\\ &-\frac2{(v-1)^3}-\frac3{(v-1)^2}-\frac3{(v-1)}\\ &+\frac{2\sqrt2}{(v-1-\sqrt2)}-\frac{2\sqrt2}{(v-1+\sqrt2)} \end{align} $$

  • $\begingroup$ 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

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.

  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.


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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