Evaluate $\int \ln(\sqrt{x+1}+\sqrt{x}) dx$

Let $x=\tan^2 \theta$. Then $dx=2\sec^2 \theta \tan \theta d\theta $. So 
\[\int \ln (\sqrt{x+1}+\sqrt{x})dx=\int \ln (\sqrt{\tan^2 \theta+1}+\sqrt{\tan^2 \theta}) 2 \sec^2 \theta \tan \theta d\theta\]
\[= \int \ln (\sec \theta +\tan \theta) 2\sec^2\theta\tan \theta d\theta.\]
Let $u=\ln (\sec \theta +\tan \theta)$ and $dv=2\sec^2\theta\tan \theta d\theta$. Then 
\[du=\frac{\sec \theta \tan \theta+\sec^2 \theta}{\sec \theta +\tan \theta}=\sec \theta,    v=\tan^2 \theta.\]
Integration by parts we get
\[\int \ln (\sec \theta +\tan \theta) 2\sec^2\theta\tan \theta d\theta\]
\[= \tan^2 \theta \ln (\sec \theta +\tan \theta)-\int tan^2 \theta \sec \theta d \theta\]
\[= \tan^2 \theta \ln (\sec \theta +\tan \theta)-\int (sec^2 \theta-1) \sec \theta d \theta\]
\[= \tan^2 \theta \ln (\sec \theta +\tan \theta)-\int sec^3 \theta d\theta+\int \sec \theta d \theta\]
\[= \tan^2 \theta \ln (\sec \theta +\tan \theta)-(\frac{1}{2}\sec \theta \tan \theta+\frac{1}{2}\ln |\sec \theta +\tan \theta|)+\ln |\sec \theta +\tan \theta|+C\]
\[= \tan^2 \theta \ln (\sec \theta +\tan \theta)-\frac{1}{2}\sec \theta \tan \theta+\frac{1}{2}\ln |\sec \theta +\tan \theta|+C\]
\[= x \ln (\sqrt{x+1} +\sqrt{x})-\frac{1}{2}\sqrt{x+1} \sqrt{x}+\frac{1}{2}\ln |\sqrt{x+1} +\sqrt{x}|+C.\]