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