$\int \sqrt{\frac{x}{x+1}}dx$ 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 than 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?
 A: 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 $$\cos \arctan x=\frac{1}{\sqrt{x^2+1}}$$
A: 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 $\sqrt{x^2-1}$.
A: Let $u=\sqrt{\frac{x}{x+1}}$. Then
$$
\begin{align}
\int\sqrt{\frac{x}{x+1}}\,\mathrm{d}x
&=\int u\,\mathrm{d}\frac{u^2}{1-u^2}\\
&=\frac12\int u\,\mathrm{d}\left(\frac1{1-u}+\frac1{1+u}\right)\\
&=\frac12\int u\left(\frac1{(1-u)^2}-\frac1{(1+u)^2}\right)\,\mathrm{d}u\\
&=\frac12\int\left(\frac1{(1-u)^2}-\frac1{1-u}+\frac1{(1+u)^2}-\frac1{1+u}\right)\,\mathrm{d}u\\
&=\frac12\left(\frac1{1-u}+\log(1-u)-\frac1{1+u}-\log(1+u)\right)+C\\
&=\frac{u}{1-u^2}+\frac12\log\left(\frac{1-u}{1+u}\right)+C\\[6pt]
&=\sqrt{x(x+1)}+\log\left(\sqrt{x+1}-\sqrt{x}\right)+C
\end{align}
$$
A: 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 :
$$\int\sqrt{\frac{x}{x+1}} \;\mathrm{d}x = \sqrt{x}\sqrt{x+1}-\operatorname{arcsinh}{\sqrt{x}} +C $$
A: $$
\begin{aligned}
\text { Let } y &=\sqrt{\frac{x}{x+1}}, \text { then } y^{2}=\frac{x}{x+1}=1-\frac{1}{x+1} \text{ and } \quad 2 y d y =\frac{1}{(x+1)^{2}} d x=\left(1-y^{2}\right)^{2} d x \\
I &=\int y \frac{2 y}{\left(1-y^{2}\right)^{2}} d y \\
&=2 \int\left(\frac{y}{1-y^{2}}\right)^{2} d y \\
&=2 \int\left[\frac{1}{2}\left(\frac{1}{1-y}-\frac{1}{1+y}\right)\right]^{2} d y \\
&=\frac{1}{2}\left[\frac{d y}{(1-y)^{2}}+\int \frac{d y}{(1+y)^{2}}-\int \frac{2}{(1-y) x(1+y} d y\right] \\
&=\frac{1}{2}\left[\frac{1}{1-y}-\frac{1}{1+y}-\int\left(\frac{1}{1-y}+\frac{1}{1+y}\right) d y\right]\\
&=\frac{\sqrt{\frac{x}{x+1}}}{1-\frac{x}{x+1}}+\ln \left|\frac{1+\sqrt{\frac{x}{x+1}}}{1-\sqrt{\frac{x}{x+1}}}\right|+C \\
&=(x+1) \sqrt{\frac{x}{x+1}}+\ln \left[\left(1+\sqrt{\frac{x}{x+1}}\right)^{2}(x+1)\right]+C\\ 
&=(x+1) \sqrt{\frac{x}{x+1}}+\ln \left|\left(1+2 \sqrt{\frac{x}{x+1}}+\frac{x}{x+1}\right)(x+1)\right|+C\mid \\
&=(x+1) \sqrt{\frac{x}{x+1}}+\ln \left|2 x+1+2(x+1) \sqrt{\frac{x}{x+1}}\right|+C
\end{aligned}
$$
$$*********$$
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$$
I=-\sqrt{x(x+1)}+\ln | 2 x+1-2 \sqrt{x(x+1)}|+C
$$
A: $\displaystyle\int\sqrt{\frac{x}{x+1}}dx$
$\displaystyle  x=y^{2}-1\Rightarrow dx=2ydy$
$\displaystyle\int\sqrt{\frac{x}{x+1}}dx=\int\frac{\sqrt{y^{2}-1}}{y}2ydy=2\int\sqrt{y^{2}-1}dy$
$\displaystyle  y=ch(z)\Rightarrow dy=sh(z)dz$
$\displaystyle sh(z)=\sqrt{ch^{2}(z)-1}=\sqrt{y^{2}-1} $
$\displaystyle\int\sqrt{\frac{x}{x+1}}dx=2\int sh^{2}(z)dz=2\int(\frac{1+ch(2z)}{2})dz=z+\frac{sh(2z)}{2}+c$
$\displaystyle\int\sqrt{\frac{x}{x+1}}dx=\ ch^{-1}{\sqrt{x+1}}+\frac{1}{2}sh\left[ 2ch^{-1}\sqrt{x+1} \right]+c$
$\displaystyle ch^{-1}(x)=ln(x+\sqrt{x^{2}-1})$
$\displaystyle \int\sqrt{\frac{x}{x+1}}dx=ln(\sqrt{x+1}+\sqrt{x})+\frac{1}{2}sh\left[ 2ln(\sqrt{x+1}+\sqrt{x}) \right]+c$
