$$ \int \frac{\sqrt{x}}{\sqrt{x}-3}dx $$

What is the most dead simple way to do this?

My professor showed us a trick for problems like this which I was able to use for the following simple example:

$$ \int \frac{1}{1+\sqrt{2x}}dx $$



being used to create


which simplifies to the answer which is:


Can I use a similar process for the first problem?

  • $\begingroup$ Hint: Long division. $\endgroup$ – Sean Roberson Sep 26 '17 at 15:57
  • $\begingroup$ First subtract and add 3 in the numerator $\endgroup$ – DanielC Sep 26 '17 at 15:57

With $\sqrt x = u$

Let $\sqrt x = u$, then we have $x = u^2$ and $\mathrm{d}x=2u\,\mathrm{d}u$.

\begin{align} \int\frac{u}{u-3}\times 2u\,\mathrm{d}u &= 2\int\frac{u^2}{u-3}\,\mathrm{d}u\\ &= 2\int\frac{u^2-9+9}{u-3}\,\mathrm{d}u\\ &= 2\int\frac{u^2-9}{u-3}\,\mathrm{d}u + 18\int\frac{1}{u-3}\,\mathrm{d}u\\ &= 2\int(u+3)\,\mathrm{d}u + 18\int\frac{1}{u-3}\,\mathrm{d}u\\ &= u^2 + 6 u+18\ln(u-3)+C_1\\ &= x+6\sqrt x +18\ln (\sqrt x -3) +C_1 \end{align}

With $\sqrt x - 3 = t$

Let $\sqrt x -3 = t$, then we have $x = (t+3)^2$ and $\mathrm{d}x=2(t+3)\,\mathrm{d}t$.

\begin{align} \int\frac{t+3}{t}\times 2(t+3)\,\mathrm{d}t &= 2\int\frac{(t+3)^2}{t}\,\mathrm{d}t \\ &= 2\int\frac{t^2+6t+9}{t}\,\mathrm{d}t \\ &= \int\left(2t+12+\frac{18}{t}\right)\,\mathrm{d}t \\ &= t^2 + 12 t + 18 \ln t +C_2\\ &= (\sqrt x-3)^2 +12(\sqrt x -2) + 18\ln(\sqrt x- 3) +C_2\\ &= x +6\sqrt x +18\ln(\sqrt x -3) + C_2-15 \end{align}

Here $C_1=C_2-15$. You can choose either way and the difference is only in the constant.

With $\frac{\sqrt x}{\sqrt x -3}=v$

It is intentionally left for the readers as an exercise.



$$\ldots = \int\frac{\sqrt{x}-3+3}{\sqrt{x}-3}\,dx = \int \left(1-\frac{3}{\sqrt{x}-3}\right)\, dx \ldots$$

  • $\begingroup$ Not really useful considering that other methods are much easier. $\endgroup$ – A---B Sep 26 '17 at 16:02
  • 2
    $\begingroup$ I think it is useful. It avoids two steps: 1) substituting and 2) "unsubstituting". OP is looking for "dead simple" way to do this. Arguably removing steps moves this in the "dead simple" direction. $\endgroup$ – Χpẘ Sep 26 '17 at 16:08
  • $\begingroup$ Moreover this technique of partial fractions is required a lot in Integration of such problems... $\endgroup$ – Aditya Sep 26 '17 at 16:52
  • $\begingroup$ @Χpẘ I don't see how you go ahead without substitution from here ? $\endgroup$ – A---B Sep 28 '17 at 15:52
  • $\begingroup$ @A---B You're right. You still need a substitution. However, I don't think other methods are "much easier" though. Artificial's answer shows that $\sqrt{x}-3=t$ is about as easy as $\sqrt{x}=u$. $\sqrt{x}-3=t$ seems a little more straightforward to me because $\int \frac{18}t dt = 18\ln{t} + C$ seems a little easier to grasp than $\int\frac{18}{u-3}du=18\ln(u-3)+C$. But that's me. $\endgroup$ – Χpẘ Sep 28 '17 at 20:03

Let $\sqrt{x}=t$.

Thus, $dx=2tdt$ and $$\int\frac{\sqrt{x}}{\sqrt{x}-3}dx=2\int\frac{t^2}{t-3}dt=2\int\frac{t^2-9+9}{t-3}dt=$$ $$=2\left(\frac{t^2}{2}+3t+9\ln|t-3|\right)+C=x+6\sqrt{x}+18\ln|\sqrt{x}-3|+C.$$

  • 2
    $\begingroup$ Or maybe easier let $t=\sqrt x -3$. $\endgroup$ – jdods Sep 26 '17 at 15:59
  • 1
    $\begingroup$ @jdods I think they are the same. $\endgroup$ – Michael Rozenberg Sep 26 '17 at 16:01
  • $\begingroup$ @MichaelRozenberg I don't think $\sqrt{x}$ and $\sqrt{x}-3$ are the same. The result'll be the same. $\endgroup$ – Χpẘ Sep 26 '17 at 16:07
  • 1
    $\begingroup$ @Χpẘ For me they are the same. I don't see any problem to find $t^2=(t-3)(t+3)+9$. $\endgroup$ – Michael Rozenberg Sep 26 '17 at 16:10
  • $\begingroup$ I think it's a little easier since just involves foiling as opposed to the +9, -9 trick. Of course, it doesn't really matter for an advanced mathematician. However a calculus student would benefit from seeing both methods. I would teach my students to look for ways to make the calculation use less work. $\endgroup$ – jdods Sep 26 '17 at 19:06

if $t=\sqrt{x}-3$, $x=\left(t+3\right)^2$ then $\dfrac{dt}{dx}=\frac{1}{2\sqrt{x}}\implies dt=\frac{dx}{2\sqrt{x}}\implies2\sqrt{x}dt=dx$ so$$\frac{\sqrt{x}}{\sqrt{x}-3}dx=\frac{\sqrt{x}}{\sqrt{x}-3}d2\sqrt{x}dt=2\frac{x}{t}dt=2\frac{\left(t+3\right)^2}{t}dt=2\left[\frac{t^2+6t}{t}+\frac{9}{t}\right]dt=2\left[t+6+\frac{9}{t}\right]dt=\left[2t+12+\frac{18}{t}\right]dt$$ integrate this:$$\int2t+12+\frac{18}{t}dt=\int2tdt+\int12dt+\int\frac{18}{t}dt=t^2+12t+18\ln t+c{=\left(\sqrt{x}-3\right)^2+12\left(\sqrt{x}-3\right)+18\ln\left|\sqrt{3}-x\right|}{=x-6\sqrt{x}+9+12\sqrt{x}-36+18\ln\left|\sqrt{3}-x\right|}\\{=x+6\sqrt{x}+18\ln\left|\sqrt{3}-x\right|-27+c,c_1=-27+c}\\\rightarrow x+6\sqrt{x}+18\ln\left|\sqrt{3}-x\right|+c_1$$


You can use the substitution $x=u^2$. So, the we differentiate both side with respect to $x$ and we get $dx=2udu$ and the the integrand becomes $\frac{2u^2}{u-3}du$. After we use it is easier to go along it.

  • $\begingroup$ You've been around awhile. Why not try your hand at using MathJax and $\LaTeX$ to improve your Answer? $\endgroup$ – hardmath May 13 '18 at 22:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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