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how to turn $ \large \int \sqrt {\frac{x}{x+1}} dx$ into a rational fraction? is this even possible?

I mean turning $\sqrt {\frac{x}{x+1}}$ into such a fraction...

Bak1139
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    Hint: use trigonometric substitution to rewrite it as a function of another variable. – Hrodelbert Jan 10 '15 at 10:35
  • I note that while five people have upvoted my answer, I am the only person to have upvoted the question so far. If you find a response to question useful enough to merit upvoting, consider upvoting the question as well! ;) – David H Jan 10 '15 at 10:54
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    @DavidH Low quality question => downvoted –  Jan 11 '15 at 16:52
  • @Fundamental have you also downvoted my other posts? cool for you... – Bak1139 Jan 12 '15 at 12:10

2 Answers2

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Hint: The obvious substitution is $\sqrt{\frac{x}{x+1}}=u$. Solving for $x$,

$$\sqrt{\frac{x}{x+1}}=u\\ \frac{x}{x+1}=u^2\\ x=u^2(x+1)=u^2x+u^2\\ (1-u^2)x=u^2\\ x=\frac{u^2}{1-u^2}.$$

Taking differentials of both sides,

$$\mathrm{d}x=\frac{2u}{(1-u^2)^2}\,\mathrm{d}u.$$

Now, the integral becomes:

$$\begin{align} \int\sqrt{\frac{x}{x+1}}\,\mathrm{d}x &=\int u\cdot \frac{2u}{(1-u^2)^2}\,\mathrm{d}u\\ &=2\int\frac{u^2}{(1-u^2)^2}\,\mathrm{d}u,\\ \end{align}$$

which is of course a rational integral.

David H
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Method-I

Let $t^2=x+1$, then: $$I=\int\frac{\sqrt{t^2-1}}{t}\cdot2t{\rm d}t=2\int\sqrt{t^2-1}{\rm d}t=t\sqrt{t^2-1}-\ln|t+\sqrt{t^2-1}|+c\\=x\sqrt{x+1}-\ln|x+\sqrt{1+x}|+c$$

Method-II

Let: $$x=\tan^2\theta\\I=\int\frac{\tan\theta\cdot2\tan\theta\sec^2\theta\rm d\theta}{\sec\theta}=2\int\tan^2\theta\sec\theta{\rm d}\theta=2\left(\int\sec^3\theta{\rm d}\theta-\int\sec\theta{\rm d}\theta\right)\\=\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|-2\ln|\sec\theta+\tan\theta|+c=x\sqrt{1+x}-\ln|x+\sqrt{1+x}|+c$$

Notes:

  • $\displaystyle\int\sqrt{x^2-a^2}{\rm d}x=\frac x2\sqrt{x^2-a^2}-\frac{a^2}2\ln|x+\sqrt{x^2-a^2}|+c$
  • $\displaystyle\int\sec^3x{\rm d}x=\frac12\sec x\tan x+\frac12\ln|\sec x+\tan x|+c$
RE60K
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