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I want to solve this integral $$\int\sqrt{\frac{x}{x+1}}dx$$ And think about:
1) $t=\frac{x}{x+1}$
2) $dt = (\frac{1}{x+1} - \frac{x}{(x+1)^2})dx$
Now I need your advice!
Thanks!

Ofir Attia
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5 Answers5

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Let $$\sqrt{\dfrac{x}{x+1}} = t \implies \dfrac{x}{x+1} = t^2 \implies x+1 = \dfrac1{1-t^2} \implies dx = \dfrac{2tdt}{(t^2-1)^2}$$ Hence, $$\int \sqrt{\dfrac{x}{x+1}} dx = \int \dfrac{2t^2}{(t^2-1)^2}dt$$ I trust you can take it from here using partial fractions. $$\dfrac{2t^2}{(t^2-1)^2} = \dfrac12 \left(\dfrac1{(t-1)^2} + \dfrac1{(t-1)} + \dfrac1{(t+1)^2} - \dfrac1{(t+1)}\right)$$

  • how you get this $$\dfrac{x}{x+1} = t^2 \implies x+1 = \dfrac1{1-t^2} \implies dx = \dfrac{2tdt}{(t^2-1)^2}$$ maybe I missed some step? – Ofir Attia May 08 '13 at 07:43
  • @OfirAttia $$\dfrac{x}{x+1} = t^2 \implies 1 - \dfrac1{x+1} = t^2 \implies \dfrac1{x+1} = 1-t^2 \implies x+1 = \dfrac1{1-t^2}$$ Hence, $$d(x+1) = d\left(\dfrac1{1-t^2} \right) \implies dx = -\dfrac{d\left(1-t^2\right)}{(1-t^2)^2} = \dfrac{2tdt}{(t^2-1)^2}$$ –  May 08 '13 at 07:43
  • After getting the equation : $$t^2=t^3(A+D)+t^2(A+B+C-D)+t(2B+2C-A-D)+(B+C-A+D)$$ I chose to find the parameters with gauss method of linear algebra and find that it dont have solution, What you are suggesting? – Ofir Attia May 08 '13 at 08:21
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put $$\sqrt{\frac{x}{x+1}}=t,\quad x=\frac{t^2}{1-t^2},\quad dx=\frac{2t}{(1-t^2)^2}\,\,dt$$

So

\begin{align*} \int\sqrt{\frac{x}{x+1}}\,\,dx &=\int t\cdot \frac{2t}{(1-t^2)^2}\,\,dt=2\int \frac{ t^2}{(1-t^2)^2}\,\,dt\\ &=2\int \frac{ t^2+1-1}{(1-t)^2(1+t)^2}\,\,dt\\ &=2\int \frac{ (t-1)(t+1)}{(1-t)^2(1+t)^2}\,\,dt-2\int \frac{1}{(1-t)^2(1+t)^2}\,\,dt\\ &=-2\int \frac{1}{ 1-t }\,\,dt-2\int \frac{1}{(1-t)^2(1+t)^2}\,\,dt\\ &=-2\ln|1-t|-2\int \frac{A}{1-t}+\frac{B}{(1-t)^2}+\frac{C}{1+t}+\frac{D}{(1+t)^2}\,\,dt\\ &=-2\ln|1-t|-2A\ln|1-t|-\frac{B}{ 1-t} -C\ln|1+t|+\frac{D}{ 1+t } \end{align*}

DeeJay
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$$\int\sqrt{\frac{x}{x+1}}\,dx$$ Let $u^2=\frac{x}{x+1}=1-\frac{1}{x+1}$. Then $2u\, du=\frac{1}{(x+1)^2}dx$. So you have $$\int2u^2(x+1)^2\,du=\int\frac{2u^2}{(1-u^2)^2}\,du$$ And it should be easy from there, as a rational function. Alternatively let $u=\sin t$ and you have $$\int\frac{2\sin^2t}{\cos^{3}t}\,dt=\int2\tan^2t\sec t\,dt$$ if a trigonometric integral is preferable. Integration by parts on this last integral yields $$2\sec t\tan t-2\int\sec^3t\,dt$$ and there is a reduction formula for integrals of powers of $\sec t$.

2'5 9'2
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If you know about Möbius transformations, from the change of variables $$ t = \frac{x}{x+1}, \quad \Longrightarrow \quad x = \frac{t}{1-t} $$ because $$ \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix}. $$ If you don't know about Möbius transformations, then you can do this : $$ t = \frac x{x+1} = 1 - \frac 1{x+1} \quad \Longrightarrow \quad 1-t = \frac 1{x+1} \quad \Longrightarrow \quad x = \frac 1{1-t} -1 = \frac t{1-t}. $$This means that $$ \left( \frac 1{x+1} - \frac x{(1+x)^2} \right) = \frac{x+1-x}{(x+1)^2} = \frac 1{(x+1)^2} = \frac{1}{\left( \frac t{1-t} + 1 \right)^2} = (1-t)^2. $$ So now you can try to solve $$ \int \frac{\sqrt t}{(1-t)^2} \, dt. $$ Letting $u = \sqrt t$, $du = \frac 1{2 \sqrt t} dt$ which implies $2u du = dt$, hence now we have $$ \int \frac{2u^2}{(1-u^2)^2} \, du $$ to solve. Use partial fractions.

Patrick Da Silva
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Allow me to make an alternate suggestion

$$x=\tan^2t,dx=2\tan t\sec^2t$$

This yields

$$\int2\tan t\sec^2t\sqrt\frac{\tan^2t}{\tan^2t+1}dt=\int2\tan t\sec^2t\times\frac{\tan t}{\sec t}dt=$$

$$\int2\tan^2t\sec tdt=2\int\sec^3tdt-2\int\sec tdt$$

I assume you've seen these integrals before. I believe the easiest way to handle the first half if I recall is to rewrite it as

$$\sec^3t=\frac1{\cos^3t}=\frac{\cos t}{\cos^4t}=\frac{\cos t}{(1-\sin^2t)^2}$$

Mike
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