Integral Of $\int\sqrt{\frac{x}{x+1}}$

Question

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

Answer 1

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. $$\dfrac{2t^2}{(t^2-1)^2} = \dfrac12 \left(\dfrac1{(t-1)^2} + \dfrac1{(t-1)} + \dfrac1{(t+1)^2} - \dfrac1{(t+1)}\right)$$

Answer 2

I’m surprised no one thought of taking $t=\sqrt{x+1}$; since the derivative of $\sqrt{x+1}$ is present in the integrand, it would be a feasible substitution valid only for $x\ge0($this is because the domain of the integrand function is $(-\infty,-1)\cup[0,\infty))$:

$$\begin{align}\int\sqrt{\frac{x}{x+1}}~\mathrm dx&=2\int\sqrt{t^2-1}~\mathrm dt\\&=t\sqrt{t^2-1}-\cosh^{-1}t+C\\&=\sqrt{x^2+x}-\sinh^{-1}\sqrt x+C\end{align}$$

For $x<-1$, we need to consider the substitution $t=\sqrt{-x-1}$:

$$\begin{align}\int\sqrt{\frac{x}{x+1}}~\mathrm dx&=-2\int\sqrt{t^2+1}~\mathrm dt\\&=-t\sqrt{t^2+1}-\sinh^{-1}t+C\\&=-\sqrt{x^2+x}-\cosh^{-1}\sqrt{-x}+C\end{align}$$

Therefore, $$\boxed{\int\sqrt{\frac{x}{x+1}}~\mathrm dx=\operatorname{sgn}(x)\sqrt{x^2+x}-\ln\left(\sqrt{|x|}+\sqrt{|x+1|}\right)+C}$$

Answer 3

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*}

Answer 4

$$\int\sqrt{\frac{x}{x+1}}\,dx$$ 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 $\sec t$.

Answer 5

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.

Answer 6

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

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

Answer 7

$$\int\sqrt{\frac x{x+1}}dx=\int\left(\frac{2x+1}{2\sqrt{x^2+x}}-\frac1{2\sqrt{x^2+x}}\right)dx =\sqrt{x^2+x}-\frac12\cosh^{-1}(2x+1)+C$$

Answer 8

Break the integral into $2$ cases for performing hyperbolic substitution:

$\textbf{Case 1, $x>0$:}$ Substitute $x=\sinh^2t, t\in[0,\infty)$:

$$\begin{align}\int\sqrt{\frac{x}{x+1}}\mathrm dx&=\int\frac{\sinh t}{\cosh t}\sinh2t\mathrm dt\\&=\int\cosh2t-1\mathrm dt\\&=\sinh t\cosh t-t+C\\&=\sqrt{x^2+x}-\sinh^{-1}\sqrt x+C\end{align}$$

$\textbf{Case 2, $x<-1$:}$ Substitute $x=-\cosh^2t, t\in[0,\infty)$:

$$\begin{align}\int\sqrt{\frac{x}{x+1}}\mathrm dx&=-\int\frac{\cosh t}{\sinh t}\sinh2t\mathrm dt\\&=\int-\cosh2t-1\mathrm dt\\&=-\sinh t\cosh t-t+C\\&=-\sqrt{x^2+x}-\cosh^{-1}\sqrt{-x}+C\end{align}$$