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

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:

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

Substituting:

$u-1=\sqrt{x}$

being used to create

$\int\frac{u-1}{u}$

which simplifies to the answer which is:

$1+\sqrt{2x}-\ln|1+\sqrt{2x}|+C$

Can I use a similar process for the first problem?

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  • $\begingroup$ 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
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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.

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Hint:

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

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  • $\begingroup$ Not really useful considering that other methods are much easier. $\endgroup$ – A---B Sep 26 '17 at 16:02
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    $\begingroup$ 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
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Let $\sqrt{x}=t$.

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

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    $\begingroup$ Or maybe easier let $t=\sqrt x -3$. $\endgroup$ – jdods Sep 26 '17 at 15:59
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    $\begingroup$ @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
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    $\begingroup$ @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
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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:$$\int2t+12+\frac{18}{t}dt=\int2tdt+\int12dt+\int\frac{18}{t}dt=t^2+12t+18\ln t+c{=\left(\sqrt{x}-3\right)^2+12\left(\sqrt{x}-3\right)+18\ln\left|\sqrt{3}-x\right|}{=x-6\sqrt{x}+9+12\sqrt{x}-36+18\ln\left|\sqrt{3}-x\right|}\\{=x+6\sqrt{x}+18\ln\left|\sqrt{3}-x\right|-27+c,c_1=-27+c}\\\rightarrow x+6\sqrt{x}+18\ln\left|\sqrt{3}-x\right|+c_1$$

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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.

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  • $\begingroup$ 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

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