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

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