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

Question

In order to integrate

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

I did

$$x = \tan^2\theta $$

$$\int \sqrt{\frac{x}{x+1}}dx = \int\sqrt{\frac{\tan^2(\theta)}{\tan^2(\theta)+1}} \ 2\tan(\theta)\sec^2(\theta)d\theta \\= \int \frac{|\tan(\theta)|}{|\sec^2(\theta)|}2\tan(\theta)\sec^2(\theta)d\theta = \int \tan^3\theta d\theta = \int\frac{\sin^3 \theta}{\cos^3 \theta}d\theta$$

$$p = \cos\theta \implies dp = -\sin\theta d\theta$$

$$\int\frac{\sin^3 \theta}{\cos^3 \theta}d\theta = -\int\frac{(1-p^2)(-\sin\theta)}{p^3 }d\theta \\= -\int \frac{1-p^2}{p^3}dp = -\int \frac{1}{p^3}dp +\int \frac{1}{p}dp \\= -\frac{p^{-2}}{-2}+\ln|p| = -\frac{(\cos\theta)^{-2}}{-2}+\ln|\cos\theta|$$

$$x = \tan^2\theta \implies \tan\theta= \sqrt{x}\implies \theta = \arctan\sqrt{x}$$

$$= -\frac{(\cos\arctan\sqrt{x})^{-2}}{-2}+\ln|\cos\arctan\sqrt{x}|$$

But the result seems a little bit different from wolfram alpha.

I Know there may be easier ways to solve this integral. But my question is about this method I choose specifically.

Is the answer correct? Also, if it is, is there a way to reduce $\cos\arctan()$ to something simpler?

Answer 1

For the last question, you have a triangle with side lengths $1,x,\sqrt{x^2+1}$ to have tangent $x$. That triangle has cosine $\frac{1}{x^2+1}$. Hence $$\cos \arctan x=\frac{1}{\sqrt{x^2+1}}$$

Answer 2

Another aproach to the integral : substitution $x=t^2$

$$\int\sqrt{\frac{x}{x+1}} \;\mathrm{d}x = \int\frac{2t^2 \mathrm{d}t}{\sqrt{t^2+1}} $$

By per partes :

$$I = \int\frac{t^2 \mathrm{d}t}{\sqrt{t^2+1}} = t\sqrt{t^2+1}-\int\sqrt{t^2+1}\;\mathrm{d}t = t\sqrt{t^2+1}-\int\frac{t^2+1}{\sqrt{t^2+1}}\;\mathrm{d}t $$

Ergo $$I = t\sqrt{t^2+1}-I-\operatorname{arcsinh}{t}$$ So, because original integral was $2I$, solution is therefore :

$$\int\sqrt{\frac{x}{x+1}} \;\mathrm{d}x = \sqrt{x}\sqrt{x+1}-\operatorname{arcsinh}{\sqrt{x}} +C $$

Answer 3

Consider $$I=\int\sqrt{\frac{x}{1+x}}dx.$$ We proced by the change of variable $u=\sqrt{1+x}$, $du=\frac{1}{2\sqrt{1+x}}$dx and $x=u^2-1,$ which gives $$I=2\int \sqrt{u^2-1}\ du=u\sqrt{u^2-1}-\log\left(u+\sqrt{u^2-1}\right)+C,$$ see https://owlcation.com/stem/How-to-Integrate-Sqrtx2-1-and-Sqrt1-x2 for more informations. The calculations ends by replacing value of u by $\sqrt{x^2-1}$.

Answer 4

Let $u=\sqrt{\frac{x}{x+1}}$. Then $$ \begin{align} \int\sqrt{\frac{x}{x+1}}\,\mathrm{d}x &=\int u\,\mathrm{d}\frac{u^2}{1-u^2}\\ &=\frac12\int u\,\mathrm{d}\left(\frac1{1-u}+\frac1{1+u}\right)\\ &=\frac12\int u\left(\frac1{(1-u)^2}-\frac1{(1+u)^2}\right)\,\mathrm{d}u\\ &=\frac12\int\left(\frac1{(1-u)^2}-\frac1{1-u}+\frac1{(1+u)^2}-\frac1{1+u}\right)\,\mathrm{d}u\\ &=\frac12\left(\frac1{1-u}+\log(1-u)-\frac1{1+u}-\log(1+u)\right)+C\\ &=\frac{u}{1-u^2}+\frac12\log\left(\frac{1-u}{1+u}\right)+C\\[6pt] &=\sqrt{x(x+1)}+\log\left(\sqrt{x+1}-\sqrt{x}\right)+C \end{align} $$

Answer 5

$$ \begin{aligned} \text { Let } y &=\sqrt{\frac{x}{x+1}}, \text { then } y^{2}=\frac{x}{x+1}=1-\frac{1}{x+1} \text{ and } \quad 2 y d y =\frac{1}{(x+1)^{2}} d x=\left(1-y^{2}\right)^{2} d x \\ I &=\int y \frac{2 y}{\left(1-y^{2}\right)^{2}} d y \\ &=2 \int\left(\frac{y}{1-y^{2}}\right)^{2} d y \\ &=2 \int\left[\frac{1}{2}\left(\frac{1}{1-y}-\frac{1}{1+y}\right)\right]^{2} d y \\ &=\frac{1}{2}\left[\frac{d y}{(1-y)^{2}}+\int \frac{d y}{(1+y)^{2}}-\int \frac{2}{(1-y) x(1+y} d y\right] \\ &=\frac{1}{2}\left[\frac{1}{1-y}-\frac{1}{1+y}-\int\left(\frac{1}{1-y}+\frac{1}{1+y}\right) d y\right]\\ &=\frac{\sqrt{\frac{x}{x+1}}}{1-\frac{x}{x+1}}+\ln \left|\frac{1+\sqrt{\frac{x}{x+1}}}{1-\sqrt{\frac{x}{x+1}}}\right|+C \\ &=(x+1) \sqrt{\frac{x}{x+1}}+\ln \left[\left(1+\sqrt{\frac{x}{x+1}}\right)^{2}(x+1)\right]+C\\ &=(x+1) \sqrt{\frac{x}{x+1}}+\ln \left|\left(1+2 \sqrt{\frac{x}{x+1}}+\frac{x}{x+1}\right)(x+1)\right|+C\mid \\ &=(x+1) \sqrt{\frac{x}{x+1}}+\ln \left|2 x+1+2(x+1) \sqrt{\frac{x}{x+1}}\right|+C \end{aligned} $$ $$*********$$

Because $\dfrac{x}{x+1}>0 \Leftrightarrow x<-1$ or $x>0$, we can simplify the answer by cases:

A. When $x\geq 0,$ as $\sqrt{(x+1)^{2}}=|x+1|=x+1$, we have $$ I=\sqrt{x(x+1)}+2 \ln (\sqrt{x}+\sqrt{x+1})+C $$ B. When $x<-1$, as $\sqrt{(x+1)^{2}}=|x+1|=-(x+1)$ , we have$$ I=-\sqrt{x(x+1)}+\ln | 2 x+1-2 \sqrt{x(x+1)}|+C $$

Answer 6

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

$\displaystyle x=y^{2}-1\Rightarrow dx=2ydy$

$\displaystyle\int\sqrt{\frac{x}{x+1}}dx=\int\frac{\sqrt{y^{2}-1}}{y}2ydy=2\int\sqrt{y^{2}-1}dy$

$\displaystyle y=ch(z)\Rightarrow dy=sh(z)dz$

$\displaystyle sh(z)=\sqrt{ch^{2}(z)-1}=\sqrt{y^{2}-1} $

$\displaystyle\int\sqrt{\frac{x}{x+1}}dx=2\int sh^{2}(z)dz=2\int(\frac{1+ch(2z)}{2})dz=z+\frac{sh(2z)}{2}+c$

$\displaystyle\int\sqrt{\frac{x}{x+1}}dx=\ ch^{-1}{\sqrt{x+1}}+\frac{1}{2}sh\left[ 2ch^{-1}\sqrt{x+1} \right]+c$

$\displaystyle ch^{-1}(x)=ln(x+\sqrt{x^{2}-1})$

$\displaystyle \int\sqrt{\frac{x}{x+1}}dx=ln(\sqrt{x+1}+\sqrt{x})+\frac{1}{2}sh\left[ 2ln(\sqrt{x+1}+\sqrt{x}) \right]+c$

Answer 7

A general approach for integrals of the form $\int f\left(x,\frac{\sqrt{x+m}}{\sqrt{x+n}}\right)\,dx$

Let $b-a=m$ and $b+a=n$, where $a$ and $b$ are real numbers. Then the following identities hold:

$$\frac{1-e^{\pm\text{i}\alpha}}{1+e^{\pm\text{i}\alpha}}=\tan{\left(\frac12\sec^{-1}\left(\frac{x+b}{a}\right)\right)}=\frac{\sqrt{x+b-a}}{\sqrt{x+b+a}}=\frac{\sqrt{x+m}}{\sqrt{x+n}},\tag{1}$$

$$e^{\pm\text{i}\alpha}=\tan\left(\frac{1}{2} \csc^{-1}\left(\frac{x+b}{a}\right)\right) = \frac{x + b - \sqrt{(x + b)^2 - a^2}}{a}\tag{2}$$

These identities are generalizations of identities $(1-2)$ and $(9-10)$ at "Integration using Euler-like identities" and can be derived similarly as shown in the linked blog. The general identities $(1)$ and $(2)$ lead us to the following general transformation formula:

$$\boxed{\int f\left(x,\tan{\frac{\beta}{2}}, \tan{\frac{\gamma}{2}} \right)\,dx=\int f\left(\frac{e^{i\alpha}+e^{-i\alpha}}{2}a, e^{\pm\text{i}\alpha}, \frac{1-e^{\pm\text{i}\alpha}}{1+e^{\pm\text{i}\alpha}}\right)\,\frac{e^{-\text{i}\alpha}-e^{\text{i}\alpha}}{2i}a\,d\alpha}\tag{3}$$ Where $\alpha=\cos^{-1}\left(\frac{x+b}{a}\right)$, $\beta=\csc^{-1}\left(\frac{x+b}{a}\right)$ and $\gamma=\sec^{-1}\left(\frac{x+b}{a}\right).$ Use the upper sign for the alternating signs $\mp$ when $\frac{x + b}{a} \geq 1$ and the lower sign when $0 \leq \frac{x + b}{a} \leq 1$.

Solution. First, we obtain the values of $a$ and $b$ by solving the following (simple!) system of equations:

$$\left. b-a=0\atop a+b=2\right\}$$

The solutions are $a = 1$ and $b = 1$. Applying formula $(3)$ for $x\geq0$, the integral becomes

$$\begin{aligned}\int \frac{\sqrt{x}}{\sqrt{x+2}}\,dx&= -\frac{i}{2}\int \frac{(1 - e^{i\alpha}) (e^{-i\alpha} - e^{i\alpha})}{(e^{i \alpha} + 1)} \, d\alpha\\&=-\frac{i}{2}\int \frac{e^{-i\alpha} (e^{i\alpha} - 1) (e^{2i\alpha} - 1)}{e^{i\alpha} + 1} \, dx\\&=-\frac{i}{2}\int e^{-i\alpha}(e^{i\alpha}-1)^2\,d\alpha\\&=-\frac{i}{2}\int (e^{-i\alpha}+e^{i\alpha}-2)\,d\alpha\\&=\frac12(2i\alpha+e^{-i\alpha}-e^{i\alpha})+C\end{aligned}$$

To switch to real numbers, replace from $(2)$, then replace the values of $a$ and $b$ and simplify, obtaining

$$=\sqrt{x^2 + 2x} + \ln\left|x+1 - \sqrt{x^2 + 2x}\right|+C$$

Now suppose you have to evaluate an integral like this one:

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

Traditional approaches can become very complicated (as you can see in the linked calculator), whereas with my method, you only need a few lines of algebra and basic calculus like in the case of your integral.